# How to find the Zeros of a Quadratic Function 4 Best methods

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.

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,

$(x - 2)^{2} = 0$

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 $x_{2} = \frac{- b - \sqrt{b^{2} - 4ac}}{2a}$

$ax^{2} + bx + c = 0$ …….. (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

#### 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

## 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.

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