A second degree polynomial equation of the form [ax^2 + bx + c =0 onumber ]where (a), (b), and (c) are any real numbers, is called a quadratic equation in (x).

The goal of this section is to develop a formulaic shortcut that will provide exact solutions of the quadratic equation ax2 +bx+c = 0. We start by moving the constant term to the other side of the equation.

[egin{array}{rlrl}{a x^{2}+b x+c} & {=0} & {} & color{Red} { ext { Quadratic equation. }} {a x^{2}+b x} & {=-c} & {} & color{Red} { ext { Subtract } c ext { from both sides. }}end{array} onumber ]

In preparation for completing the square, we next divide both sides of the equation by (a).

[x^{2}+dfrac{b}{a} x=-dfrac{c}{a} quad ext { Divide both sides by } a onumber ]

Now we complete the square. Take one-half of the coefficient of (x), then square the result.

(dfrac{1}{2} cdot dfrac{b}{a}=dfrac{b}{2 a}) when squared gives (left(dfrac{b}{2 a} ight)^{2}=dfrac{b^{2}}{4 a^{2}})

We now add (dfrac{b^{2}}{4 a^{2}}) to both sides of the equation.

[x^{2}+dfrac{b}{a} x+dfrac{b^{2}}{4 a^{2}}=-dfrac{c}{a}+dfrac{b^{2}}{4 a^{2}} quad color {Red} ext { Add } b^{2} /left(4 a^{2} ight) ext { to both sides. } onumber ]

On the left, we factor the perfect square trinomial. On the right, we make equivalent fractions with a common denominator.

[egin{array}{ll}{left(x+dfrac{b}{2 a} ight)^{2}=-dfrac{c}{a} cdot dfrac{4 a}{4 a}+dfrac{b^{2}}{4 a^{2}}} & color {Red} { ext { On the left, factor. On the right, }} {} & color {Red} { ext { create equivalent fractions with }} {left(x+dfrac{b}{2 a} ight)^{2}=-dfrac{4 a c}{4 a^{2}}+dfrac{b^{2}}{4 a^{2}}} & color {Red} { ext { Multiply numerators and denominators. }} {left(x+dfrac{b}{2 a} ight)^{2}=dfrac{b^{2}-4 a c}{4 a^{2}}} & color {Red} { ext { Add fractions. }}end{array} onumber ]

When we take the square root, there are two answers.

[x+dfrac{b}{2 a}=pm sqrt{dfrac{b^{2}-4 a c}{4 a^{2}}} quad color {Red} ext { Two square roots. } onumber ]

When you take the square root of a fraction, you take the square root of both the numerator and denominator.

[egin{aligned} x+dfrac{b}{2 a} &=pm dfrac{sqrt{b^{2}-4 a c}}{sqrt{4 a^{2}}} x+dfrac{b}{2 a} &=pm dfrac{sqrt{b^{2}-4 a c}}{2 a} quad color {Red} ext { Simplify: } sqrt{4 a^{2}}=2 a x &=-dfrac{b}{2 a} pm dfrac{sqrt{b^{2}-4 a c}}{2 a} quad color {Red} ext { Subtract } b /(2 a) ext { from both sides } end{aligned} onumber ]

Because both fractions have the same denominator, we can add and subtract numerators and put the answer over the common denominator.

[x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a} onumber ]

The equation (a x^{2}+b x+c=0) is called a quadratic equation. Its solutions are given by[x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a} onumber ]called the quadratic formula.

Whew! Fortunately, the result is a lot easier to apply than it is to develop! Let’s try some examples.

Example (PageIndex{1})

Solve for (x: x^{2}-4 x-5=0)

Solution

The integer pair (1,−5) has product (ac = −5) and sum (b = −4). Hence, this trinomial factors.

[egin{array}{r}{x^{2}-4 x-5=0} {(x+1)(x-5)=0}end{array} onumber ]

Now we can use the zero product property to write:

[egin{array}{rlrl}{x+1} & {=0} & { ext { or }} & {x-5} & {=0} {x} & {=-1} & {} & {x} & {=5}end{array} onumber ]

Thus, the solutions are (x =−1) and (x = 5). Now, let’s give the quadratic formula a try. First, we must compare our equation with the quadratic equation, then determine the values of (a), (b), and (c).

[egin{array}{l}{a x^{2}+b x+c=0} {x^{2}-4 x-5=0}end{array} onumber ]

Comparing equations, we see that (a = 1), (b = −4), and (c = −5). We will now plug these numbers into the quadratic formula. First, replace each occurrence of (a), (b), and (c) in the quadratic formula with open parentheses.

[egin{aligned}
x &=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a} quad color {Red} ext { The quadratic formula. } x& = {dfrac{-(quad) pm sqrt{(quad)^{2}-4( )( )}}{2( )} quad quad color {Red} ext { Replace } a, b, ext { and } c ext { with open parentheses. }}end{aligned} onumber ]

Now we can substitute: (1) for (a), (−4) for (b), and (−5) for(c).

[egin{array}{ll}{x=dfrac{-(-4) pm sqrt{(-4)^{2}-4(1)(-5)}}{2(1)}} & color {Red} { ext { Substitute: } 1 ext { for } a,-4 ext { for } b} {x=dfrac{4 pm sqrt{16+20}}{2}} & color {Red} { ext { Simplify. Exponent first, then }} {x=dfrac{4 pm sqrt{36}}{2}} & color {Red} { ext { Add: } 16+20=36} {x=dfrac{4 pm 6}{2}} & color {Red} { ext { Simplify: } sqrt{36}=6}end{array} onumber ]

Note that because of the “plus or minus” symbol, we have two answers.

[egin{array}{ll}{x=dfrac{4-6}{2}} & ext {or} & {x=dfrac{4+6}{2}} {x=dfrac{-2}{2}} && {x=dfrac{10}{2}} {x=-1} && {x=5}end{array} onumber ]

Note that these answers match the answers found using the ac-test to factor the trinomial.

Exercise (PageIndex{1})

Solve for (x: x^{2}-8x+12=0)

(2), (6)

Example (PageIndex{2})

Solve for (x : x^{2}=5 x+7)

Solution

The equation is nonlinear, make one side zero.

[egin{array}{rlrl}{x^{2}} & {=5 x+7} & {} & color {Red} { ext { Original equation. }} {x^{2}-5 x-7} & {=0} & {} & color {Red} { ext { Nonlinear. Make one side zero. }}end{array} onumber ]

Compare (x^2 −5x−7 = 0) with (ax^2 + bx + c = 0) and note that (a = 1), (b = −5), and (c = −7). Replace each occurrence of (a), (b), and (c) with open parentheses to prepare the quadratic formula for substitution.

[egin{array}{ll}{x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a}} & color {Red} { ext { The quadratic formula. }} {x=dfrac{-( ) pm sqrt{( )^{2}-4( )( )}}{2( )}} & color {Red} { ext { Replace } a, b, ext { and } c ext { with }}end{array} onumber ]

Substitute (1) for (a), (−5) for (b), and (−7) for (c).

[egin{array}{ll}{x=dfrac{-(-5) pm sqrt{(-5)^{2}-4(1)(-7)}}{2(1)}} & color {Red} { ext { Substitute: } a=1, b=-5, c=-7} {x=dfrac{5 pm sqrt{25+28}}{2}} & color {Red} { ext { Exponents and multiplication first. }} {x=dfrac{5 pm sqrt{53}}{2}} & color {Red} { ext { Simplify. }}end{array} onumber ]

Check: Use the calculator to check each solution (see Figure (PageIndex{1})). Note that in storing ((5-sqrt{53}) / 2) in (mathbf{X}), we must surround the numerator in parentheses.

Figure (PageIndex{1}): Check ((5-sqrt{53}) / 2) and ((5+sqrt{53}) / 2).

In each image in Figure (PageIndex{1}), after storing the solution in (mathbf{X}), note that the left and right-hand sides of the original equation (x^2 =5 x + 7) produce the same number, verifying that our solutions are correct.

Exercise (PageIndex{2})

Solve for (x : x^{2}+7 x=10)

((-7+sqrt{89}) / 2,(-7-sqrt{89}) / 2)

Example (PageIndex{3})

Solve for (x : 7 x^{2}-10 x+1=0)

Solution

Compare (7x^2 −10x + 1 = 0) with (ax^2 + bx + c = 0) and note that (a = 7), (b = −10), and (c = 1). Replace each occurrence of (a), (b), and (c) with open parentheses to prepare the quadratic formula for substitution.

[egin{aligned}
x &=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a} quad color {Red} ext { The quadratic formula. }
x &= dfrac{-(quad) pm sqrt{( )^{2}-4( )( )}}{2( )} quad color {Red} ext { Replace } a, b, ext { and } c ext { with open parentheses.}
end{aligned} onumber ]

Substitute (7) for (a), (−10) for (b), and (1) for (c).

[egin{array}{ll}{x=dfrac{-(-10) pm sqrt{(-10)^{2}-4(7)(1)}}{2(7)}} & color {Red} { ext { Substitute: } 7 ext { for } a} {x=dfrac{10 pm sqrt{100-28}}{14}} & color {Red} { ext { Exponent, then multiplication. }} {x=dfrac{10 pm sqrt{72}}{14}} & color {Red} { ext { Simplify. }}end{array} onumber ]

In this case, note that we can factor out a perfect square, namely (sqrt{36}).

[egin{array}{ll}{x=dfrac{10 pm sqrt{36} sqrt{2}}{14}} & color {Red} {sqrt{72}=sqrt{36} sqrt{2}} {x=dfrac{10 pm 6 sqrt{2}}{14}} & color {Red} { ext { Simplify: } sqrt{36}=6}end{array} onumber ]

Finally, notice that both numerator and denominator are divisible by (2).

[egin{aligned}
x&= dfrac{ frac{10 pm 6 sqrt{2}}{2}}{ frac{14}{2}} quad color {Red} ext { Divide numerator and denominator by } 2.
x&= dfrac{ frac{10}{2} pm frac{6 sqrt{2}}{2}}{ frac{14}{2}} quad color {Red} ext { Distribute the } 2. x&=dfrac{5 pm 3 sqrt{2}}{7} quad color {Red} ext { Simplify. }
end{aligned} onumber ]

Alternate simplification: Rather than dividing numerator and denominator by (2), some prefer to factor and cancel, as follows.

[egin{array}{ll}{x=dfrac{10 pm 6 sqrt{2}}{14}} & color {Red} { ext { Original answer. }} {x=dfrac{2(5 pm 3 sqrt{2})}{2(7)}} & color {Red} { ext { Factor out a } 2} {x=dfrac{ ot{2}(5 pm 3 sqrt{2})}{ ot{2}(7)}} & color {Red} { ext { Cancel. }} {x=dfrac{5 pm 3 sqrt{2}}{7}} & color {Red} { ext { Simplify. }}end{array} onumber ]

Note that we get the same answer using this technique.

Exercise (PageIndex{3})

Solve for (x : 3 x^{2}+8 x+2=0)

((-4+sqrt{10}) / 3,(-4-sqrt{10}) / 3)

Example (PageIndex{4})

An object is launched vertically and its height (y) (in feet) above ground level is given by the equation (y = 320+192t−16t^2), wheret is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level? After placing the answer in simple form and reducing, use your calculator to round the answer to the nearest tenth of a second.

Solution

When the object returns to ground level, its height (y) above ground level is (y = 0) feet. To find the time when this occurs, substitute (y = 0) in the formula (y = 320 + 192t−16t^2) and solve for (t).

[egin{array}{ll}{y=320+192 t-16 t^{2}} & color {Red} { ext { Original equation. }} {0=320+192 t-16 t^{2}} & color {Red} { ext { Set } y=0}end{array} onumber ]

Each of the coefficients is divisible by (−16).

[0=t^{2}-12 t-20 quad color{Red} ext { Divide both sides by }-16 onumber ]

Compare (t^2−12t−20 = 0) with (at^2 +bt+c = 0) and note that (a = 1), (b = −12), and (c = −20). Replace each occurrence of (a), (b), and (c) with open parentheses to prepare the quadratic formula for substitution. Note that we are solving for t this time, not (x).

[egin{array}{ll}{x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a}} & color {Red} { ext { The quadratic formula. }} {x=dfrac{-( ) pm sqrt{( )^{2}-4( )( )}}{2( )}} & color {Red} { ext { Replace } a, b, ext { and } c ext { with open parentheses. }}end{array} onumber ]

Substitute (1) for (a), (−12) for (b), and (−20) for (c).

[egin{array}{ll}{t=dfrac{-(-12) pm sqrt{(-12)^{2}-4(1)(-20)}}{2(1)}} & color {Red} { ext { Substitute: } 1 ext { for } a} {t=dfrac{12 pm sqrt{144+80}}{2}} & color {Red} { ext { Exponent, then multiplication. }} {t=dfrac{12 pm sqrt{224}}{2}} & color {Red} { ext { Simplify. }}end{array} onumber ]

The answer is not in simple form, as we can factor out (sqrt{16}).

[egin{array}{ll}{t=dfrac{12 pm sqrt{16} sqrt{14}}{2}} & color {Red} {sqrt{224}=sqrt{16} sqrt{14}} {t=dfrac{12 pm 4 sqrt{14}}{2}} & color {Red} { ext { Simplify: } sqrt{16}=4}end{array} onumber ]

Use the distributive property to divide both terms in the numerator by (2).

[egin{array}{ll}{t=dfrac{12}{2} pm dfrac{4 sqrt{14}}{2}} & color{Red} { ext { Divide both terms by } 2} {t=6 pm 2 sqrt{14}} & color {Red} { ext { Simplify }}end{array} onumber ]

Thus, we have two solutions, (t=6-2 sqrt{14}) and (t=6+2 sqrt{14}). Use your calculator to find decimal approximations, then round to the nearest tenth.

Figure (PageIndex{2}): Using calculator to find decimal approximations

[t approx-1.5,13.5 onumber ]

The negative time is irrelevant, so to the nearest tenth of a second, it takes the object approximately (13.5) seconds to return to ground level.

Exercise (PageIndex{4})

An object is launched vertically and its height (y) (in feet) above ground level is given by the equation (y = 160 + 96t−16t^2), where (t) is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level?

(3+sqrt{19} approx 7.4) seconds

Example (PageIndex{5})

Arnie gets on his bike at noon and begins to ride due north at a constant rate of (12) miles per hour. At 1:00 PM, Barbara gets on her bike at the same starting point and begins to ride due east at a constant rate of (8) miles per hour. At what time of the day will they be (50) miles apart (as the crow ﬂies)? Don’t worry about simple form, just report the time of day, correct to the nearest minute.

Solution

At the moment they are (50) miles apart, let (t) represent the time that Arnie has been riding since noon. Because Barbara started at 1:00 PM, she has been riding for one hour less than Arnie. So, let (t−1) represent the numbers of hours that Barbara has been riding at the moment they are (50) miles apart.

Now, if Arnie has been riding at a constant rate of (12) miles per hour for (t) hours, then he has traveled a distance of (12t) miles. Because Barbara has been riding at a constant rate of (8) miles per hour for (t−1) hours, she has traveled a distance of (8(t−1)) miles.

Figure (PageIndex{3}): (50) miles apart.

The distance and direction traveled by Arnie and Barbara are marked in Figure (PageIndex{3}). Note that we have a right triangle, so the sides of the triangle must satisfy the Pythagorean Theorem. That is,

[(12 t)^{2}+[8(t-1)]^{2}=50^{2} quad color{Red} ext { Use the Pythagorean Theorem. } onumber ]

Distribute the (8).

[(12 t)^{2}+(8 t-8)^{2}=50^{2} quad color{Red} ext { Distribute the } 8 onumber ]

Square each term. Use ((a−b)^2 = a^2 −2ab + b^2) to expand ((8t−8)^2).

[egin{aligned} 144 t^{2}+64 t^{2}-128 t+64 &=2500 quad color{Red} ext { Square each term. } 208 t^{2}-128 t+64 &=2500 quad color{Red} ext { Simplify: } 144 t^{2}+64 t^{2}=208 t^{2} end{aligned} onumber ]

The resulting equation is nonlinear. Make one side equal to zero.

[egin{array}{rlrl}{208 t^{2}-128 t-2436} & {=0} & {} & color{Red} { ext { Subtract } 2500 ext { from both sides. }} {52 t^{2}-32 t-609} & {=0} & {} & color{Red} { ext { Divide both sides by } 4 .}end{array} onumber ]

Compare (52t^2 −32t−609 = 0) with (at^2 + bt + c = 0) and note that (a = 52), (b = −32), and (c = −609). Note that we are solving for (t) this time, not (x).

[egin{array}{ll}{x=dfrac{-b pm sqrt{b^{2}-4 a c}}{2 a}} & color{Red} { ext { The quadratic formula. }} {x=dfrac{-( ) pm sqrt{( )^{2}-4( )( )}}{2( )}} & color{Red} { ext { Replace } a, b, ext { and } c ext { with }}end{array} onumber ]

Substitute (52) for (a), (−32) for (b), and (−609) for (c).

[egin{align*} t &= dfrac{-(-32) pm sqrt{(-32)^{2}-4(52)(-609)}}{2(52)} quad color {Red} ext { Substitute: } 52 ext { for } a t &= dfrac{32 pm sqrt{1024+126672}}{104} quad color {Red} ext {Exponent, then multiplication.} t &= dfrac{32 pm sqrt{127696}}{104} quad color {Red} ext { Simplify. } end{align*} onumber ]
Now, as the request is for an approximate time, we won’t bother with simple form and reduction, but proceed immediately to the calculator to approximate this last result (see Figure (PageIndex{4})). Thus, Arnie has been riding for approximately (3.743709336) hours. To change the fractional part (0.743709336) hours to minutes, multiply by (60)min/hr.

Figure (PageIndex{4}): Approximate time that Arnie has been riding.

[0.743709336 mathrm{hr}=0.743709336 mathrm{hr} imes dfrac{60 mathrm{min}}{mathrm{hr}}=44.62256016 mathrm{min} onumber ]

Rounding to the nearest minute, Arnie has been riding for approximately (3) hours and (45) minutes. Because Arnie started riding at noon, the time at which he and Barbara are (50) miles apart is approximately 3:45 PM.

Exercise (PageIndex{5})

At 6:00 AM, a freight train passes through Sagebrush Junction heading west at (40) miles per hour. At 8:00 AM, a passenger train passes through the junction heading south at (60) miles per hour. At what time of the day, correct to the nearest minute, will the two trains be (180) miles apart?

9:42 AM

The quadratic formula is a formula used to solve quadratic equations. It is the solution to the general quadratic equation. Quadratics are polynomials whose highest power term has a degree of 2 .

a, b and c are constants, where a cannot equal 0 . The ± indicates that the quadratic formula has two solutions. Each of these is referred to as a root. Geometrically, these roots represent the points at which a parabola crosses the x-axis . Thus, the quadratic formula can be used to determine the zeros of any parabola, as well as give the axis of symmetry of the parabola.

If a quadratic is missing either the bx or c term, then set b or c equal to 0 . If the quadratic does not contain the ax 2 term, you cannot use the quadratic formula because the denominator of the quadratic formula will equal 0 . In that case, you can use algebra to find the zeros.

The quadratic formula mainly involves plugging numbers into the equation, but there are a few things you need to know. The part of the formula within the radical is called the discriminant:

The discriminant tells us how many solutions the quadratic has.

In addition, notice the ± symbol. This means that when the discriminant is positive, the quadratic will have two solutions - one where you add the square root of the discriminant, and one where you subtract it.

Below is an example of using the quadratic formula:

Although the quadratic equation may at first seem daunting to remember, repeated use can help. If you know the tune to "Pop goes the weasel," you can also sing the quadratic equation to its tune to help you remember the quadratic equation. The song goes:

"x is equal to negative b , plus or minus the square root, of b squared minus 4ac all over 2a ."

Label vertex point, use axis of sym.

Observation:
a > 1, then parabola is _________
a < 1, then parabola is _________
a < 0, then parabola is _________

The Graph of f (x) = a(x – h) 2

Label vertex point, use axis of sym.

Observation:
(x+h) h is neg, then parabola _________
(x–h) h is pos, then parabola _________
The effect of h is _________

The Graph of f (x) = a(x – h) 2 + k

Label vertex point, use axis of sym.

Observation:
The vertex is _________
The equation of the A.O.S. _________
The effect of k _________

Graphing y = a(x – h) 2 + k

Graph:

Summarize the Vertex Form

y = a(x – h) 2 + k is the Vertex Form.

Need To Know
▪ Review completing the square
▪ Converting into Vertex Form
(not to learn but to appreciate the short cut formula)
▪ Vertex Point Formula – Short Cut
▪ Finding intercepts
▪ Sketching the graph of a quadratic

Completing the Square

y = a(x – h) 2 + k
What is the vertex point? ____________

If the quadratic is in standard form, we have no information.
We need to change the form into vertex form (squared stuff).

g(x) = x 2 – 10x + 21
f(x) = 4x 2 + 8x - 3

Vertex Point – short cut (easy way)

If f(x) = ax 2 + bx + c, then (h, k) = ___________
which means _____________________________.
1. Find h with the formula
2. Find k by plugging h into the function.

g(x) = x 2 – 10x + 21
f(x) = 4x 2 + 8x – 3

Recall:
X-intercept point = point where graph crosses x-axis. (a, 0)
Find the x-intercept point by letting y = 0.

Y-intercept point = point where graph crosses y-axis. (0, b)
Find the y-intercept point by letting x = 0.
Example:

## Ms. McCullough's Math Class

i did number 25 on our homework but i do not think i got it right. can you tell me were i went wrong?

im not sure if you can understand this or not. sorry

what do i do if the "a" in the problem does not have and exponet?

Hi Davis! About your first comment. you set it up correctly, but remember if you have a negative number inside your square root, it means that there are no real solutions (because we can't take the square root of a negative number). About your second comment. if "a" does not have an exponent of 2, then it is not really "a". You will have to rearrange the terms so that it goes in order (ax^2+bx+c=0), but which problem are you referring to?

I was waiting on someone to catch that. it's a trick question. It's not quadratic, so you can just solve for x and you will only have one answer.

oh thanks Ms. McCullough thats nice of you. so all i have to do is solve for x?

Plugging in the values of a , b , and c , you will get the desired values of x .

If the expression under the square root sign ( b 2 − 4 a c , also called the discriminant) is negative, then there are no real solutions. (You need complex numbers to deal with this case properly. These are usually taught in Algebra 2 .)

If the discriminant is zero, there is only one solution. If the discriminant is positive, then the ± symbol means you get two answers.

Here a = 1 , b = − 1 , and c = − 12 . Substituting, we get:

x = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 12 ) 2 ( 1 )

The discriminant is positive, so we have two solutions:

In this example, the discriminant was 49 , a perfect square, so we ended up with rational answers. Often, when using the quadratic formula, you end up with answers which still contain radicals.

Here a = 3 , b = 2 , and c = 1 . Substituting, we get:

x = − 2 ± 2 2 − 4 ( 3 ) ( 1 ) 2 ( 3 )

The discriminant is negative, so this equation has no real solutions.

Here a = 1 , b = − 4 , and c = 2 . Substituting, we get:

x = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = 4 ± 16 − 8 2 = 4 ± 8 2

x = 4 ± 4 ⋅ 2 2 = 4 ± 2 2 2 = 2 ( 2 ± 2 ) 2 = 2 ± 2

The discriminant is positive but not a perfect square, so we have two real solutions:

## Applied Algebra: Modeling and functions

Not every quadratic equation can be solved by factoring or by extraction of roots. For example, the expression (x^2 + x - 1) cannot be factored, so the equation (x^2 + x - 1 = 0) cannot be solved by factoring. For other equations, factoring may be difficult. In this section we learn two methods that can be used to solve any quadratic equation.

Instead of completing the square every time we solve a new quadratic equation, we can complete the square on the general quadratic equation,

and obtain a formula for the solutions of any quadratic equation.

The solutions of the equation (ax^2 + bx + c = 0 ext

This formula expresses the solutions of a quadratic equation in terms of its coefficients. (The proof of the formula is considered in the Homework problems.) The symbol (pm ext<,>) read plus or minus, is used to combine the two equations

To solve a quadratic equation using the quadratic formula, all we have to do is substitute the coefficients (a ext<,>) (b ext<,>) and (c) into the formula.

###### Example 6.18 .

Write the equation in standard form as

Using a calculator, we find that the solutions are approximately (1.707) and (0.293 ext<.>)

We can also verify that the (x)-intercepts of the graph of (y = 2x^2 - 4x + 1) are approximately (1.707) and (0.293 ext<,>) as shown below.

###### Checkpoint 6.19 .

Use the quadratic formula to solve (

### Subsection Applications

We have now seen four different algebraic methods for solving quadratic equations:

1. Factoring
2. Extraction of roots
3. Completing the square

Factoring and extraction of roots are relatively fast and simple, but they do not work on all quadratic equations. The quadratic formula will work on any quadratic equation.

###### Example 6.20 .

The owners of a day-care center plan to enclose a divided play area against the back wall of their building, as shown below. They have (300) feet of picket fence and would like the total area of the playground to be (6000) square feet. Can they enclose the playground with the fence they have, and if so, what should the dimensions of the playground be?

Suppose the width of the play area is (x) feet. Because there are three sections of fence along the width of the play area, that leaves (300 - 3x) feet of fence for its length. The area of the play area should be (6000) square feet, so we have the equation

This is a quadratic equation. In standard form,

The left side cannot be factored, so we use the quadratic formula with (a = alert<1> ext<,>) (b = alert<-100> ext<,>) and (c = alert<2000> ext<.>)

Simplifying the last fraction, we find that (x approx 72.35) or (xapprox 27.65 ext<.>) Both values give solutions to the problem.

• If the width of the play area is (72.35) feet, then the length is (300 - 3(72.35) ext<,>) or (82.95) feet.
• If the width is (27.65) feet, the length is (300 - 3(27.65) ext<,>) or (217.05) feet.
###### Checkpoint 6.21 .

In Investigation 6.1, we considered the height of a baseball, given by the equation

Find two times when the ball is at a height of (20) feet. Give your answers to two decimal places.

Sometimes it is useful to solve a quadratic equation for one variable in terms of the others.

###### Example 6.22 .

We first write the equation in standard form as a quadratic equation in the variable (x ext<.>)

Expressions in (y) are treated as constants with respect to (x ext<,>) so that (a = alert<1> ext<,>) (b = alert<-y> ext<,>) and (c = alert ext<.>) Substitute these expressions into the quadratic formula.

### Subsection Introduction to complex numbers

You know that not all quadratic equations have real solutions.

has no (x)-intercepts (as shown at right), and the equation

We can still use completing the square or the quadratic formula to solve the equation.

###### Example 6.24 .

Solve the equation (x^2 - 2x + 2 = 0) by using the quadratic formula.

We substitute (a = 1 ext<,>) (b =-2 ext<,>) and (c= 2) into the quadratic formula to get

Because (sqrt<-4>) is not a real number, the equation (x^2 - 2x + 2 = 0) has no real solutions.

###### Checkpoint 6.25 .

Solve the equation (x^2 - 6x + 13 = 0) by using the quadratic formula.

### Subsubsection Imaginary Numbers

Although square roots of negative numbers such as (sqrt<-4>) are not real numbers, they occur often in mathematics and its applications.

Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. René Descartes gave them the name imaginary numbers, which reflected the mistrust with which mathematicians regarded them at the time. Today, however, such numbers are well understood and used routinely by scientists and engineers.

We begin by defining a new number, (i ext<,>) whose square is (-1 ext<.>)

###### Caution 6.26 .

The letter (i) used in this way is not a variable it is the name of a specific number and hence is a constant.

The square root of any negative number can be written as the product of a real number and (i ext<.>) For example,

or (sqrt<-4>=2i ext<.>) Any number that is the product of (i) and a real number is called an .

###### Imaginary Numbers.

Examples of imaginary numbers are

###### Example 6.27 .

Write each radical as an imaginary number.

1. (displaystyle egin[t] sqrt<-25>amp=sqrt<-1>sqrt<25> amp=isqrt<25>=5i end)
2. (displaystyle egin[t] 2sqrt<-3>amp=2sqrt<-1>sqrt<3> amp=2isqrt <3>end)
###### Checkpoint 6.28 .

Write each radical as an imaginary number.

###### Note 6.29 .

Every negative real number has two imaginary square roots, (isqrt) and (-isqrt ext<,>) because

For example, the two square roots of (-9) are (3i) and (-3i ext<.>)

### Subsubsection Complex Numbers

Using the quadratic formula to solve the equation, we find

If we now replace (sqrt<-16>) with (4i ext<,>) we have

The two solutions are (1 + 2i) and (1 - 2i ext<.>) These numbers are examples of .

###### Complex Numbers.

A can be written in the form (a+bi ext<,>) where (a) and (b) are real numbers.

Examples of complex numbers are

In a complex number (a+bi ext<,>) (a) is called the , and (b) is called the . All real numbers are also complex numbers (with the imaginary part equal to zero). A complex number whose real part equals zero is called a number.

###### Example 6.30 .

Write the solutions to Example 6.24, (dfrac<2pmsqrt<-4>><2> ext<,>) as complex numbers.

Because (sqrt<-4>=sqrt<-1>sqrt<4>=2i ext<,>) we have (dfrac<2pmsqrt<-4>><2>=dfrac<2pm2i><2> ext<,>) or (1pm i ext<.>) The solutions are (1+i) and (1-i ext<.>)

###### Checkpoint 6.31 .

Use extraction of roots to solve ((2x + 1)^2 + 9 = 0 ext<.>) Write your answers as complex numbers.

### Subsection Arithmetic of Complex Numbers

All the properties of real numbers listed in Algebra Skills Refresher Section A.13 are also true of complex numbers. We can carry out arithmetic operations with complex numbers.

We add and subtract complex numbers by combining their real and imaginary parts separately. For example,

###### Example 6.32 .

Subtract: ((8 - 6i ) - (5 + 2i ) ext<.>)

Combine the real and imaginary parts.

###### Checkpoint 6.33 .

Subtract: ((-3 + 2i ) - (-3 - 2i ) ext<.>)

### Subsection Section Summary

#### Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.

#### Subsubsection CONCEPTS

The square of the binomial is a ,

###### To Solve a Quadratic Equation by Completing the Square.

Write the equation in standard form.

Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides.

Complete the square on the left side:

1. Multiply the coefficient of the first-degree term by one-half, then square the result.
2. Add the value obtained in (a) to both sides of the equation.

Write the left side of the equation as the square of a binomial. Simplify the right side.

Use extraction of roots to finish the solution.

The solutions of the equation (ax^2 + bx + c = 0 ext

We have four methods for solving quadratic equations: extracting of roots, factoring, completing the square, and using the quadratic formula. The first two methods are faster, but they don't work on all equations. The last two methods work on any quadratic equation.

#### Subsubsection STUDY QUESTIONS

Name four algebraic methods for solving a quadratic equation.

Give an example of a quadratic trinomial that is the square of a binomial.

What number must be added to (x^2 - 26x) to make it the square of a binomial?

After completing the square, how do we finish solving the quadratic equation?

What is the first step in solving the equation (2x^2 - 6x = 5) by completing the square?

#### Subsubsection SKILLS

Practice each skill in the Homework problems listed.

Solve quadratic equations by completing the square: #3–24

Solve problems by writing and solving quadratic equations: #37–44

### Exercises Homework 6.2

For Problems 1-2, complete the square and write the result as the square of a binomial.

For Problems 3-18, solve by completing the square.

For Problems 19-24, solve by completing the square. Your answers will involve (a ext<,>) (b ext<,>) or (c ext<.>)

Write an expression for the area of the square in the figure.

Express the area as a polynomial.

Divide the square into four pieces whose areas are given by the terms of your answer to part (b).

Write an expression for the area of the shaded region in the figure.

Express the area in factored form.

By making one cut in the shaded region, rearrange the pieces into a rectangle whose area is given by your answer to part (b).

A car traveling at (s) miles per hour on a dry road surface requires approximately (d) feet to stop, where (d) is given by the function

Make a table showing the stopping distance, (d ext<,>) for speeds of (10 ext<,>) (20 ext<,>) (ldots) , (100) miles per hour. (Use the feature of your calculator.)

Graph the function for (d) in terms of (s ext<.>) Use your table values to help you choose appropriate window settings.

Write and solve an equation to answer the question: If a car must be able to stop in (50) feet, what is the maximum safe speed it can travel?

A car traveling at (s) miles per hour on a wet road surface requires approximately (d) feet to stop, where (d) is given by the function

Make a table showing the stopping distance, (d ext<,>) for speeds of (10 ext<,>) (20 ext<,>) (ldots) , (100) miles per hour. (Use the feature of your calculator.)

Graph the function for (d) in terms of (s ext<.>) Use your table values to help you choose appropriate window settings.

Insurance investigators at the scene of an accident find skid marks (100) feet long leading up to the point of impact. Write and solve an equation to discover how fast the car was traveling when it put on the brakes. Verify your answer on your graph.

A skydiver jumps out of an airplane at (11,000) feet. While she is in free-fall, her altitude in feet (t) seconds after jumping is given by the function

Make a table of values showing the skydiver's altitude at (5)-second intervals after she jumps from the airplane. (Use the feature of your calculator.)

Graph the function. Use your table of values to choose appropriate window settings.

If the skydiver must open her parachute at an altitude of (1000) feet, how long can she free-fall? Write and solve an equation to find the answer.

If the skydiver drops a marker just before she opens her parachute, how long will it take the marker to hit the ground? (Hint: The marker continues to fall according to the equation given above.)

A high diver jumps from the (10)-meter springboard. His height in meters above the water (t) seconds after leaving the board is given by the function

Make a table of values showing the diver's altitude at (0.25)-second intervals after he jumps from the airplane. (Use the feature of your calculator.)

Graph the function. Use your table of values to choose appropriate window settings.

How long is it before the diver passes the board on the way down?

How long is it before the diver hits the water?

A dog trainer has (100) meters of chain link fence. She wants to enclose (250) square meters in three pens of equal size, as shown in the figure.

Let (l) and (w) represent the length and width, respectively, of the entire area. Write an equation about the amount of chain link fence.

Solve your equation for (l) in terms (w ext<.>)

Write and solve an equation in (w) for the total area enclosed.

Find the dimensions of each pen.

An architect is planning to include a rectangular window topped by a semicircle in his plans for a new house, as shown in the figure. In order to admit enough light, the window should have an area of (120) square feet. The architect wants the rectangular portion of the window to be (2) feet wider than it is tall.

Let (x) stand for the horizontal width of the window. Write expressions for the height of the rectangular portion and for the radius of the semicircular portion.

Write an expression for the total area of the window.

Write and solve an equation to find the width and overall height of the window.

When you look down from a height, say a tall building or a mountain peak, your line of sight is tangent to the Earth at the horizon, as shown in the figure.

Suppose you are standing on top of the Petronas Tower in Kuala Lumpur, (1483) feet high. How far can you see on a clear day? (You will need to use the Pythagorean theorem and the fact that the radius of the Earth is (3960) miles. Do not forget to convert the height of the Petronas Tower to miles.)

How tall a building should you stand on in order to see (100) miles?

If the radius of the Earth is (6370) kilometers, how far can you see from an airplane at an altitude of (10,000) meters? (Hint: See Problem 43.)

b. How high would the airplane have to be in order for you to see a distance of (10) kilometers?

For Problems 45-52, use the quadratic formula to solve each equation for the indicated variable.

In order to sketch the graph of the quadratic equation, we follow these steps :

(a) Check if a > 0 or a < 0 to decide if it is U-shaped or n-shaped.

(b) The Vertex: The x-coordinate of the minimum point (or maximum point) is given by

(which can be shown using completing the square method, which we met earlier).

We substitute this x-value into our quadratic function (the y expression). Then we will have the (x, y) coordinates of the minimum (or maximum) point. This is called the vertex of the parabola.

(c) The coordinates of the y-intercept (substitute x = 0). This is always easy to find!

(d) The coordinates of the x-intercepts (substitute y = 0 and solve the quadratic equation), as long as they are easy to find.

### Example 1

Sketch the graph of the function y = 2x^2&minus 8x + 6

We first identify that a = 2, b = -8 and c = 6.

### Step (a)

Since a = 2, a > 0 hence the function is a parabola with a minimum point and it opens upwards (U-shaped)

### Step (b)

The x co-ordinate of the minimum point is:

The y value of the minimum point is

So the minimum point is (2, -2)

### Step (c)

The y-intercept is found by substituting x = 0 into the y expression.

So (0, 6) is the y-intercept.

### Step (d)

The x-intercepts are found by setting y = 0 and solving:

2x^2 - 8x + 6 = 0

2(x^2 - 4x + 3) = 0

2(x - 1)(x - 3) = 0

So x = 1, or x = 3.

The given quadratic equation is in the form of

Substitute the above values of a, b and c into the quadratic formula.

Therefore, the solution is

The given quadratic equation is in the form of

Substitute the above values of a, b and c into the quadratic formula.

Therefore, the solution is

The given quadratic equation is in the form of

Substitute the above values of a, b and c into the quadratic formula.

Therefore, the solution is

Write the given quadratic equation in the form :

Substitute the above values of a, b and c into the quadratic formula.

Therefore, the solution is

Write the given quadratic equation in the form :

Substitute the above values of a, b and c into the quadratic formula.

Therefore, the solution is

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## News from Mathnasium of Pflugerville

For those students in high school, and even some younger, we are familiar with the quadratic formula, "the opposite of b, plus or minus the square root of b squared minus 4ac, all divided by 2a". This formula allows you to find the root of quadratic equations of the form: ax 2 + bx + c = 0.

Where did this formula come from? Why did older civilizations need to solve equations of this form in the first place? The following article, taken from h2g2, explores the origins of this famous formula.

Original article: https://h2g2.com/approved_entry/A2982567

This is the quadratic formula, as it is taught to most of us in school:

x1,2=(-b/2a) ± (1/2a)(b 2 -4ac) 1/2

gives the solution to a generic quadratic equation of the form:

ax 2 + bx + c = 0

The development, or derivation, of a mathematical idea is usually as logical, deducible and rectilinear as possible. This brings about the common notion that its historical development is similarly as continuous, logical and rectilinear: one mathematician picking up an idea where another mathematician left it.

Using the quadratic formula as an example, it will be shown that the historical development of mathematics is not at all rectilinear. Instead, parallel developments, interconnections and confluences can be found, which - to complicate this stuff even further - are also interrelated with social, cultural, political and religious matters.

The so-called quadratic formula has been derived in the course of a few millennia to its current form, which is taught to most of us in school. This Entry will strictly concentrate on the historical development of the quadratic formula. Some mathematical background may be of use to fully understand the described development, however the maths used in this Entry will be kept at a necessary minimum.

### The Original Problem 2000(or so)BC

Egyptian, Chinese and Babylonian engineers were really smart people - they knew how the area of a square scales with the length of its side. They knew that it's possible to store nine times more bales of hay if the side of the square loft is tripled. They also found out how to calculate the area of more complex designs like rectangles and T-shapes and so on. However, they didn't know how to calculate the sides of the shapes - the length of the sides, starting from a given area - which was often what their clients really needed. And so, this is the original problem: a certain shape 1 must be scaled with a total area, and in the end what's needed is lengths of the sides, or walls to make a working floor plan.

### 1500BC The Beginnings - Egypt

The first aspect that finally led to the quadratic equation was the recognition that it is connected to a very pragmatic problem, which in its turn demanded a 'quick and dirty' solution. We have to note, in this context, that Egyptian mathematics did not know equations and numbers like we do nowadays it is instead descriptive, rhetorical and sometimes very hard to follow. It is known that the Egyptian wisemen (engineers, scribes and priests) were aware of this shortcoming - but they came up with a way to circumvent this problem: instead of learning an operation, or a formula that could calculate the sides from the area, they calculated the area for all possible sides and shapes of squares and rectangles and made a look-up table. This method works much like we learn the multiplication tables by heart in school instead of doing the operation proper.

So, if someone wanted a loft with a certain shape and a certain capacity to store bales of papyrus, the engineer would go to his table and find the most fitting design. The engineers did not have time to calculate all shapes and sides to make their own table. Instead, the table they used was a reproduction of a master look-up table. The copyists did not know if the stuff they were copying made sense or not as they didn't know anything about maths. So, obviously, sometimes errors crept in, and copies of the copies were known to be less trustworthy 2 . These tables still exist, and it is possible to see where errors crept in during the copying of the documents.

### 400 BCE The Next Step - Babylon and China

The Egyptian method worked fine, but a more general solution - without the need for tables - seemed desirable. That's where the Babylonian geeks come into play. Babylonian maths had a big advantage over the one used in Egypt, namely they used a number-system that is pretty much like the one we use today, albeit on a hexagesimal basis, or base-60. Addition and multiplication were a lot easier to perform with this system, so the engineers around 1000 BC could always double-check the values in their tables. By 400 BC they found a more general method called 'completing the square' to solve generic problems involving areas. There are no indications that these people used a specific mathematical procedure to find out the solutions, so probably some educated guessing was involved. Around the same time, or a bit later, this method also appears in Chinese documents. The Chinese, like the Egyptians, also did not use a numeric system, but a double checking of simple mathematical operations was made astonishingly easy by the widespread use of the abacus.

### 300BC Geometry - Hellenistic Mediterranean Area

The first attempts to find a more general formula to solve quadratic equations can be tracked back to geometry (and trigonometry) top-bananas Pythagoras (500 BC in Croton, Italy) and Euclid (300 BC in Alexandria, Egypt), who used a strictly geometric approach, and found a general procedure to solve the quadratic equation. Pythagoras noted that the ratios between the area of a square and the respective length of the side - the square root - were not always integer, but he refused to allow for proportions other than rational. Euclid went even further and found out that this proportion might also not be rational. He concluded that irrational numbersexist.

Euclid's opus Elements covered more or less all the mathematics needed for technical applications from a theoretical point of view. However, it didn't use the same notation with formulas and numbers like we use nowadays. For that reason it was not possible to calculate the square root of any number by hand, in order to obtain a good approximation for the exact value of the root, which is what the architects and engineers were after. Because all (theoretically relevant at least) maths seemed to be complete 3 but otherwise useless, the many wars occurring in Europe, and also the early Middle Ages turned the mathematical world in Europe silent until the 13th Century. In this period mathematics also suffered a big shift, going from a pragmatic science to a more mystical, philosophical discipline.

### 700AD All Numbers - India

Hindu mathematics has used the decimal system (the one we use) at least since 600AD. One of the most important influences on Hindu mathematics was that it was widely used in commerce. The average Hindu merchant was pretty fast in simple maths. If someone had a debt the numbers would be negative, if someone had a credit the numbers would be positive. Also, if someone had neither credit, nor debt, the numbers would add up to zero. Zero is an important number in the history of mathematics, and its relatively late appearance is due to the fact that many cultures had difficulty of conceiving 'nothing'. The concept of 'nothing', like in 'shunya', the void, or the concept of 'equilibrium', was already anchored in Hindu culture.

Around 700AD the general solution for the quadratic equation, this time using numbers, was devised by a Hindu mathematician called Brahmagupta, who, among other things, used irrational numbers he also recognised two roots in the solution. The final, complete solution as we know it today came around 1100AD, by another Hindu mathematician called Baskhara 4 . Baskhara was the first to recognise that any positive number has two square roots.

### 820AD Powerful Islamic Science - Persia

Around 820AD, near Baghdad, Mohammad bin Musa Al-Khwarismi, a famous Islamic mathematician 5 who knew Hindu mathematics, also derived the quadratic equation. The algebra used by him was entirely rhetorical, and he rejected negative solutions. This particular derivation of the quadratic formula was brought to Europe by Jewish mathematician/astronomer Abraham bar Hiyya (whose Latinised name is Savasorda) who lived in Barcelona around 1100.

With the Renaissance in Europe, academic attention came back to original mathematical problems. By 1545 Girolamo Cardano, who was a typical Renaissance scientist (ie, interested in alchemy, occultism and suchlike), and one of the best algebraists of his time, compiled the works related to the quadratic equations - that is, he blended Al-Khwarismi's solution with the Euclidean geometry. He was possibly not the first or only one, but the most famous. In his (mainly rhetorical) works he allows for the existence of complex, or imaginary numbers - that is, roots of negative numbers. At the end of the 16th Century the mathematical notation and symbolism was introduced by amateur-mathematician François Viète, in France. In 1637, when René Descartes published La Géométrie, modern Mathematics was born, and the quadratic formula has adopted the form we know today.

## Frequently asked questions on finding the zeros of a quadratic function

### How many zeros can a quadratic function have?

A quadratic function has 2 zeros real or complex.

### How many real zeros can a quadratic function have?

A quadratic function has either 2 real zeros or 0 real zeros.
We know that complex roots occur in conjugate pairs.
Therefore a quadratic function can not have one complex root ( or zero).

### What are the zeros of the quadratic function f(x) = 8x^2 – 16x – 15?

Given quadratic function is f(x) = 8x^ <2>- 16x - 15 .
Comparing this with the quadratic function ax^ <2>+ bx + c = 0 , we get
a = 18, b = - 16, c = -15
Now putting these values of a, b, c on Quadratic formula we get
x = frac <- b pm sqrt- 4ac>> <2a>
or, x = frac <- (-16) pm sqrt<(-16)^<2>- 4(8)(-15)>> <2(8)>
or, x = frac< 16 pm sqrt<256 + 480>> <16>
or, x = frac< 16 pm sqrt<736>> <16>
or, x = frac< 16 pm 4sqrt<46>> <16>
or, x = frac< 4 pm sqrt<46>> <4>
or, x = frac< 4 + sqrt<46>><4>,frac< 4 - sqrt<46>> <4>
Therefore the zeros of the quadratic function f(x) = 8ࡨ – 16x – 15 are x = frac< 4 + sqrt<46>><4>, frac<4 - sqrt<46>> <4>.

### Which is a zero of the quadratic function f(x) = 16x^2 + 32x − 9?

Given quadratic function is f(x) = 16x^ <2>+ 32x - 9 .
We will find the zeros of the quadratic function f(x) = 16x^ <2>+ 32x - 9 by factoring.
16x^ <2>+ 32x - 9 = 0
or, 16x^ <2>+ (36 - 4)x - 9 = 0
or, 16x^ <2>+ 36x - 4x - 9 = 0
or, 4x (4x + 9) -1 (4x + 9) = 0
or, (4x + 9)(4x -1) = 0
Either 4x + 9 = 0 or 4x - 1 = 0
Either 4x = -9 or 4x = 1
Either x = frac<-9> <4>or x = frac<1> <4>
Therefore the zeros of the quadratic function f(x) = 16x^ <2>+ 32x - 9 are x = frac<-9><4>, : frac<1> <4>.

### What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

Given quadratic function is f(x) = 6x^ <2>+ 12x – 7 .
We will find the zeros of the quadratic function by the quadratic formula.
Comparing this with the quadratic function ax^ <2>+ bx + c = 0 , we get
a = 6, b = 12, c = -7
Now putting these values of a, b, c on Quadratic formula we get
x = frac <- b pm sqrt- 4ac>> <2a>
or, x = frac <- 12 pm sqrt<(12)^<2>- 4(6)(-7)>> <2(6)>
or, x = frac<- 12 pm sqrt<144 + 168>> <12>
or, x = frac<- 12 pm sqrt<312>> <12>
or, x = frac<- 12 pm 2 sqrt<78>> <12>
or, x = frac<- 6 pm sqrt<78>> <6>
or, x = frac<- 6 + sqrt<78>><6>, frac<- 6 - sqrt<78>> <6>
Therefore the zeros of the quadratic function f(x) = 6x^ <2>+ 12x – 7 are x = frac<- 6 + sqrt<78>><6>, : frac<- 6 - sqrt<78>> <6>.

### What are the zeros of the quadratic function f(x) = 2x^2 + 16x – 9?

Given quadratic function is f(x) = 2x^ <2>+ 16x – 9 .
We use the quadratic formula to find the zeros of the quadratic function f(x) = 2x^ <2>+ 16x – 9 .
Comparing this with the quadratic function ax^ <2>+ bx + c = 0 , we get
a = 2, b = 16, c = -9
Now putting these values of a, b, c on Quadratic formula we get
x = frac <- b pm sqrt- 4ac>> <2a>
or, x = frac <- 16 pm sqrt<(16)^<2>- 4(2)(-9)>> <2(2)>
or, x = frac<- 16 pm sqrt<256 + 72>> <4>
or, x = frac<- 16 pm sqrt<328>> <4>
or, x = frac<- 16 pm 2 sqrt<82>> <4>
or, x = frac<- 8 pm sqrt<82>> <2>
or, x = frac<- 8 + sqrt<82>><2>, frac<- 8 - sqrt<82>> <2>
Therefore the zeros of the quadratic function f(x) = 2x^ <2>+ 16x – 9 are x = frac<- 8 + sqrt<82>><2>, : frac<- 8 - sqrt<82>> <2>.

### The zeros of a quadratic polynomial are 1 and 2 then what is the polynomial?

The quadratic polynomial whose zeros are 1 and 2 is
(x-1)(x-2)
= x(x-2) -1(x-2)
= x^ <2>- 2x -x +2
= x^ <2>-3x + 2

### What are the zeroes of the quadratic polynomial 3x^2-48?

We can write
3x^<2>-48=0
or, 3(x^<2>-16)=0
or, x^<2>-16=0 (Dividing both sides by 3)
or, x^<2>=16
or, x=pm sqrt <16>
or, x=pm 4
Therefore the zeroes of the quadratic polynomial 3x^2-48 are x = +4, -4.

### 3x+1/x-8=0 is a quadratic equation or not

We know that the degree of a quadratic function is 2.
But the degree of the function frac<3x+1> is not equal to 2.
Therefore the given function frac<3x+1> is not a quadratic function.
Consequently, 3x+1/x-8=0 is not a quadratic equation.

### Find quadratic polynomial whose sum of roots is 0 and the product of roots is 1.

Let the roots of the quadratic polynomial are ‘a’ and ‘b’.
Then by the given condition, we have,
a+b=0
or, a=-b
and
ab=1
or, (-b)b=1
or, b^<2>=-1
or, b=pm sqrt <-1>
or, b=+sqrt<-1>, -sqrt <-1>
Now a=-b=- (sqrt<-1>) = mp sqrt <-1>=-sqrt<-1>, +sqrt <-1>
If we take a=-sqrt <-1>and b=+sqrt <-1>then the quadratic polynomial is
(x-a)(x-b)
= (x-(-sqrt<-1>))(x-sqrt<-1>)
= (x+sqrt<-1>)(x-sqrt<-1>)
= (x)^<2>-(sqrt<-1>)^ <2>
= x^<2>-(-1)
= x^<2>+1
Again if we take a=+sqrt <-1>and b=-sqrt <-1>then the quadratic polynomial is
(x-a)(x-b)
= (x-sqrt<-1>)(x-(-sqrt<-1>))
= (x-sqrt<-1>)(x+sqrt<-1>)
= (x)^<2>-(sqrt<-1>)^ <2>
= x^<2>-(-1)
= x^<2>+1
Therefore the quadratic polynomial whose sum of roots (zeros) is 0 and the product of roots (zeros) is 1 is x^<2>+1 and the zeros of the quadratic polynomial are x= +sqrt<-1>, -sqrt <-1>.

We hope you understand how to find the zeros of a quadratic function.

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