how to find the zeros of a quadratic function

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?

Table of Contents - What you will learn

• What is a quadratic function
• Quadratic function in standard form
• Quadratic function examples
• How to find the zeros of a quadratic function – 4 best methods
  • Find zeros of a quadratic function by Completing the square
  • How to find zeros of a quadratic function by Factoring
  • Finding zeros of a function using Quadratic formula
    • Proof of Quadratic formula
  • How to find zeros of a Quadratic function on a graph
    • Find zero when the graph cut the x-axis
    • When the graph touches the x-axis
    • When the graph neither touch nor cut the x-axis
• Frequently asked questions on finding the zeros of a quadratic function
  • How many zeros can a quadratic function have?
  • How many real zeros can a quadratic function have?
  • What are the zeros of the quadratic function f(x) = 8x^2 – 16x – 15?
  • Which is a zero of the quadratic function f(x) = 16x^2 + 32x − 9?
  • What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?
  • What are the zeros of the quadratic function f(x) = 2x^2 + 16x – 9?
  • The zeros of a quadratic polynomial are 1 and 2 then what is the polynomial?
  • What are the zeroes of the quadratic polynomial 3x^2-48?
  • 3x+1/x-8=0 is a quadratic equation or not
  • Find quadratic polynomial whose sum of roots is 0 and the product of roots is 1.

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

• Completing the square,
• Factoring,
• Quadratic formula,
• Graphing.

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} = 0

or, 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} = 0

i.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 = 0

or, \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 = 0

or, - \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)(x-5) = 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} =0

or, 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 x-axis 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

1. When the graph cut the x-axis,
2. When the graph touches the x-axis,
3. When the graph neither touch nor cut the x-axis.

Find zero when the graph cut the x-axis

Look at the graph of the function \left ( x+2 \right )^{2}=4\left ( y+4 \right ) given below

Here the graph cut the x-axis 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 x-axis 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 x-axis

Look at the graph of the function \left ( x-1 \right )^{2}=4y given below

Here the graph does not cut the x-axis 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 x-axis.

But the graph of the quadratic function y = x^{2} touches the x-axis 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 x-axis

Look at the graph of the function x^{2}=4\left ( y-2 \right ) given below

Here the graph neither cut nor touch the x-axis.

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 x-axis.

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 = 0

or, 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

How to find the zeros of a Quadratic function on a graph
Source – Youtube, Video by Marty Brandl

Frequently asked questions on finding the zeros of a quadratic function

1. How many zeros can a quadratic function have?

   A quadratic function has 2 zeros real or complex.

2. 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).

3. 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} .

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} .

5. 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} .

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} .

7. 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

8. 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.

9. 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}{x-8} is not equal to 2.
   Therefore the given function \frac{3x+1}{x-8} is not a quadratic function.
   Consequently, 3x+1/x-8=0 is not a quadratic equation.

10. 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.

If you have any doubts or suggestions on the topic of how to find the zeros of a quadratic function feel free to ask in the comment section. We love to hear from you.

Additionally, you can read:

• What is a function? – Definition, Example, and Graph.
• 48 Different Types of Functions and there Examples and Graph – [Complete list].
• How to find the zeros of a function – 3 Best methods

how to find the zeros of a quadratic function

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