How to find the zeros of a quadratic function?
In the previous lesson, we have discussed how to find the zeros of a function.
Now we will know 4 best methods of finding the zeros of a quadratic function.
But before that, we have to know what is a quadratic function?
What is a quadratic function
A quadratic function is a polynomial function of degree 2.
Quadratic function in standard form
The standard form of a quadratic function is
y=ax^{2}+bx+c,
where a, b, c are constants.
Quadratic function examples
Some examples of quadratic function are
 y = x^{2} ,
 y = 3x^{2}  2x ,
 y = 8x^{2}  16x  15 ,
 y = 16x^{2} + 32x  9 ,
 y = 6x^{2} + 12x  7 ,
 y = \left ( x  2 \right )^{2} .
How to find the zeros of a quadratic function – 4 best methods
There are different methods of finding the zeros of a quadratic function.
We will see the best 4 methods of them
Find zeros of a quadratic function by Completing the square
There are some quadratic polynomial functions of which we can find zeros by making it a perfect square.
This is the easiest way to find the zeros of a polynomial function.
For example, y = x^{2}  4x + 4 is a quadratic function. We can easily convert it into a square using the formula \left ( a  b \right )^{2} = a^{2} 2ab + b^{2} like this
x^{2}  4x + 4= (x)^{2}  2\times 2\times x + (2)^{2}
= (x  2)^{2} ,
which is a perfect square.
Now the next step is to equate this perfect square with zero and get the zeros (roots) the given quadratic function.
Equating with zero we get,
(x  2)^{2} = 0or, x = 2, 2.
There the zeros of the quadratic function y = x^{2}  4x + 4 are x = 2, 2.
Here 2 is a root of multiplicity 2.
We will see two more examples to understand the concept completely.
Question: How do you find the zeros of a quadratic function \frac{z^{2}}{4} + \frac{5z}{3} + \frac{25}{9} by using the method of completing the square?
Answer: First we make the given quadratic a perfect square and then equate the square with zero.
\frac{z^{2}}{4} + \frac{5z}{3} + \frac{25}{9}= \left ( \frac{z}{2} \right )^{2} + 2 \times \frac{z}{2} \times \frac{5}{3} + \left ( \frac{5}{3} \right )^{2}
= \left ( \frac{z}{2} + \frac{5}{3} \right )^{2} ( by using a^{2} + 2ab + b^{2} = \left ( a + b \right )^{2} )
Equating with zero, we get
\left ( \frac{z}{2} + \frac{5}{3} \right )^{2} = 0i.e., \left ( \frac{z}{2} + \frac{5}{3} \right ) = 0 and \left ( \frac{z}{2} + \frac{5}{3} \right ) = 0
i.e., \frac{z}{2} = \frac{5}{3} and \frac{z}{2} = \frac{5}{3}
i.e., z = \frac{10}{3} and z = \frac{10}{3}
Therefore the roots of a quadratic function \frac{z^{2}}{4} + \frac{5z}{3} + \frac{25}{9} are z = \frac{10}{3}, \frac{10}{3} .
Here \frac{10}{3} is a root of multiplicity 2.
Question: How to find the zeros of a quadratic function y = 49x^{2}  42x + 9 by using the method of completing the square
Answer: We find the zeros of the quadratic function y = 49x^{2}  42x + 9 like the previous example.
49x^{2}  42x + 9 = 0or, \left ( 7x \right )^{2}  2\times 7x\times 3 + \left ( 3 \right )^{2} =0
or, \left ( 7x  3 \right )^{2} =0
or, 7x  3 = 0 and 7x  3 = 0
or, 7x = 3 and 7x = 3
or, x = \frac{3}{7} and x = \frac{3}{7}
Therefore the zeros of a quadratic function y = 49x^{2}  42x + 9 are x = \frac{3}{7}, \frac{3}{7}
How to find zeros of a quadratic function by Factoring
In this method, we have to find the factors of the given quadratic function.
For example, x^{2}  x  6 is a quadratic function and we have to find the zeros of this function.
For this purpose, we find the factors of this function.
First, we multiply the coefficient of x^{2} i.e., 1 with 6
coefficient of x^{2}\times 6 = 1 \times 6 = 6
In the given function the sign of the coefficient of x^{2} is positive and the sign of 6 is negative.
Next, we have to find two factors of 6 such that the difference between the factors of 6 will give 1 as the coefficient of x is 1.
Two such factors of 6 are 3 and 2 and the difference is 3 – 2 = 1.
Next, follow the steps as given below
x^{2}  x  6 = 0= x^{2}  (3  2)x  6 = 0
= x^{2}  3x  2x  6 = 0
= x (x  3)  2 (x  3) = 0
= (x  3)(x  2) = 0
Either x  3 = 0 or x  2 = 0
Either x = 3 or x = 2
Therefore the zeros of a quadratic function x^{2}  x  6 are 3 and 2.
Now look at the two examples given below
Question: How do you find the zeros of a quadratic function  x^{2}  3x + 40 .
Answer: The given quadratic function is  x^{2}  3x + 40 .
Here the coefficient of x^{2} is 1 which is negative.
In factor method of finding the zeros of a quadratic function, we need the sign of the leading term x^{2} to be positive.
For that reason first, we take common – 1 from the quadratic function
 x^{2}  3x + 40 = 0or,  \left ( x^{2} + 3x  40 \right ) = 0
After that, we repeat the process shown in the previous example like this
or,  \left ( x^{2} + 3x  40 \right ) = 0
or, – { x^{2} + (8  5)x  40 } = 0 ( Since 8 and 5 are two factors of 40 and 8 – 5 = 3)
or,  \left ( x^{2} + 8x  5x  40 \right ) = 0
or,  \left ( x^{2} + 8x  5x  40 \right ) = 0
or,  \left ( x(x + 8)  5(x + 8 \right ) = 0
or, – { x(x + 8)  5(x + 8 } = 0
or, (x+8)(x5) = 0
Either x + 8 = 0 or x  5 = 0
Either x =  8 or x = 5
Therefore the zeros of a quadratic function  x^{2}  3x + 40 are x =  8, 5 .
Question: How to find the zeros of a quadratic function x^{2}  \frac{5x}{6} + \frac{1}{6}
Answer: Product of the coefficient of x^{2} and the constant term \frac{1}{6} is \frac{1}{6} and the sign of the constant term \frac{1}{6} is positive.
So we have to find two factors of \frac{1}{6} such that the sum of these factors will be \frac{5}{6} i.e. the coefficient of x .
Such two factors of \frac{1}{6} are \frac{1}{2} and \frac{1}{3} and their sum = \frac{1}{2} + \frac{1}{3} = \frac{5}{3} .
Now the solution is
x^{2}  \frac{5x}{6} + \frac{1}{6} =0or, x^{2}  \left ( \frac{1}{2} + \frac{1}{3} \right )x + \frac{1}{6} =0
or, x^{2}  \frac{1}{2}x  \frac{1}{3}x + \frac{1}{6} =0
or, x \left ( x  \frac{1}{2} \right )  \frac{1}{3} \left ( x  \frac{1}{2} \right ) = 0
or, \left ( x  \frac{1}{2} \right ) \left ( x  \frac{1}{3} \right ) = 0
Either x  \frac{1}{2} = 0 or x  \frac{1}{3} = 0
Either x = \frac{1}{2} or x = \frac{1}{3}
Therefore the zeros of a quadratic function x^{2}  \frac{5x}{6} + \frac{1}{6} are x = \frac{1}{2}, \frac{1}{3}
Finding zeros of a function using Quadratic formula
The Quadratic formula is a formula for finding the zeros of a quadratic function.
Let ax^{2} + bx +c = 0 be a quadratic function where a, b, c are constants with a \neq 0 , then the quadratic formula is
x = \frac{ b \pm \sqrt{b^{2}  4ac}}{2a} ,
where ” \pm ” shows that the quadratic function has two zeros.
i.e., if x_{1} and x_{2} be two zeros of the quadratic function ax^{2} + bx +c = 0 , then
x_{1} = \frac{ b + \sqrt{b^{2}  4ac}}{2a} and x_{2} = \frac{ b  \sqrt{b^{2}  4ac}}{2a}
Proof of Quadratic formula
ax^{2} + bx + c = 0 …….. (1)
or, 4a^{2}x^{2} + 4abx + 4ac = 0 (Multiplying bothsides by 4a)
or, 4a^{2}x^{2} + 4abx =  4ac
or, 4a^{2}x^{2} + 4abx +b^{2} = b^{2}  4ac ( by adding b^{2} on bothsides)
or, (2ax)^{2} + 2 \times 2ax \times b +(b)^{2} = b^{2}  4ac
or, \left ( 2ax + b \right )^{2} = b^{2}  4ac ( by using x^{2} +2xy + y^{2} = \left ( x + y \right )^{2} )
or, 2ax + b = \pm \sqrt{ b^{2}  4ac}
or, 2ax =  b \pm \sqrt{ b^{2}  4ac}
or, x =\frac{  b \pm \sqrt{ b^{2}  4ac}}{2a} ……. (2)
or, x = \frac{  b + \sqrt{ b^{2}  4ac}}{2a}, \: \frac{  b  \sqrt{ b^{2}  4ac}}{2a} ……. (3)
Now we find the zeros of some quadratic function using Quadratic formula:
Question: How to find the zeros of a quadratic function x^{2}  x  6 = 0
Answer: Given that x^{2}  x  6 = 0 and we have to find the zeros of this quadratic function.
Comparing this with the quadratic function ax^{2} + bx + c = 0 , we get
a = 1, b = 1, and c= 6.
Now putting these values in equation (3) we get
x = \frac{  (1) + \sqrt{ (1)^{2}  4\times (1) \times (6)}}{2 \times 1}, \frac{  (1)  \sqrt{ (1)^{2}  4 \times (1) \times (6)}}{2 \times 1}or, x = \frac{ 1 + \sqrt{ 1 + 24}}{2}, \frac{ 1  \sqrt{ 1 + 24}}{2}
or, x = \frac{ 1 + \sqrt{25}}{2}, \frac{ 1  \sqrt{25}}{2}
or, x = \frac{ 1 + 5 }{2}, \frac{ 1  5 }{2}
or, x = \frac{ 6 }{2}, \frac{  4 }{2}
or, x = 3,  2
Therefore the zeros of a quadratic function x^{2}  x  6 = 0 are x = 3,  2 .
Question: How do you find the zeros of a quadratic function x^{2} + 1
Answer: Given that x^{2} + 1 = 0 .
We can write this function as x^{2} + 0 \times x + 1 = 0
We will find the zeros of this quadratic function using the Quadratic formula.
Comparing this with the quadratic function ax^{2} + bx + c = 0 , we get
a = 1, b = 0, c = 1.
Now putting these values of a, b, c in equation (3) we get
x = \frac{ b + \sqrt{b^{2}  4ac}}{2a}, \frac{ b  \sqrt{b^{2}  4ac}}{2a}or, x = \frac{ 0 + \sqrt{(0)^{2}  4 (1)(1)}}{2 (1)}, \frac{ 0  \sqrt{(0)^{2}  4 (1)(1)}}{2 (1)}
or, x = \frac{+ \sqrt{4}}{2}, \frac{ \sqrt{4}}{2}
or, x = \frac{+ 2 \sqrt{1}}{2}, \frac{2 \sqrt{1}}{2}
or, x = + \sqrt{1},  \sqrt{1}
or, x = + i,  i
Therefore the zeros of the quadratic function x^{2} + 1 = 0 are x = + i,  i and both of them are complex (not real).
How to find zeros of a Quadratic function on a graph
To find the zero on a graph what we have to do is look to see where the graph of the function cut or touch the xaxis and these points will be the zero of that function because at these point y is equal to zero.
Here 3 cases will arise and they are
 When the graph cut the xaxis,
 When the graph touches the xaxis,
 When the graph neither touch nor cut the xaxis.
Find zero when the graph cut the xaxis
Look at the graph of the function \left ( x+2 \right )^{2}=4\left ( y+4 \right ) given below
Here the graph cut the xaxis at two points (6,0) and (2,0).
At (6,0), x=6; y=0 and at (2,0), x=2; y=0.
We can clearly see that the function value y=0 for x=6 and 2.
There the zeros of the function are 6 and 2.
Question: How do you find the zeros of a quadratic function on the graph y = x^{2}  2
To find the zeros of the quadratic function y = x^{2}  2 on the graph first we have to plot the quadratic function y = x^{2}  2 on the graph.
From the graph, we can see that the quadratic function y = x^{2}  2 cuts the xaxis at x = 1.4 and x = 1.4 .
So the quadratic function y = x^{2}  2 has two real zeros and they are x = 1.4 and x = 1.4
Also as the degree of a quadratic function is 2 and the number of roots (or zeros) of a quadratic function is 2, therefore x^{2}  2 has no complex zeros.
When the graph touches the xaxis
Look at the graph of the function \left ( x1 \right )^{2}=4y given below
Here the graph does not cut the xaxis but touch at (1,0).
At (1,0), x=1 and y=0.
Clearly the function value y=0 for x=1.
There the zero of the function is 1.
Question: How do you find the zeros of a quadratic function on the graph y = x^{2}
Look at the graph of the quadratic function y = x^{2} .
Here we can clearly see that the quadratic function y = x^{2} does not cut the xaxis.
But the graph of the quadratic function y = x^{2} touches the xaxis at point C (0,0).
Therefore the zero of the quadratic function y = x^{2} is x = 0.
Now you may think that y = x^{2} has one zero which is x = 0 and we know that a quadratic function has 2 zeros.
Actually, the zero x = 0 is of multiplicity 2.
What I mean to say that the zeros of the quadratic function y = x^{2} are x = 0, 0 and they are real.
When the graph neither touch nor cut the xaxis
Look at the graph of the function x^{2}=4\left ( y2 \right ) given below
Here the graph neither cut nor touch the xaxis.
So we have no real value of x for which y=0.
In this case, we have no real zero of the function.
Question: How do you find the zeros of a quadratic function on the graph y = x^{2} + 2
Look at the graph of the quadratic function y = x^{2} + 2 given on the right side.
Here you can clearly see that the graph of y = x^{2} + 2 neither cut nor touch the xaxis.
Therefore the function y = x^{2} + 2 has no real zeros.
If we solve the equation y = x^{2} + 2 = 0 we will found two complex zeros of y = x^{2} + 2 = 0
x^{2} + 2 = 0or, x^{2} =  2
or, x = \pm \sqrt{ 2}
or, x = \pm \sqrt{2} i
or, x = + \sqrt{2} i, \sqrt{2} i
For better understanding, you can watch this video (duration: 5 min 29 sec) where Marty Brandl explained the process for finding zeros on a graph
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{b^{2}  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 4\sqrt{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) = 8×2 – 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{b^{2}  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{b^{2}  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
(x1)(x2)
= x(x2) 1(x2)
= x^{2}  2x x +2
= x^{2} 3x + 2 
What are the zeroes of the quadratic polynomial 3x^248?
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^248 are x = +4, 4. 
3x+1/x8=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}{x8} is not equal to 2.
Therefore the given function \frac{3x+1}{x8} is not a quadratic function.
Consequently, 3x+1/x8=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
(xa)(xb)
= (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
(xa)(xb)
= (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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