### #ezw_tco-2 .ez-toc-widget-container ul.ez-toc-list li.active{ background-color: #ededed; } chapter outline

The roots of a quadratic equation are the values of the variable that satisfies the equation. They are also known as the ‘zeroes’ of the quadratic equation. For the equation ax2 + bx + c = 0 the two roots α and β are:

• ${\alpha =\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}}$
• β = ${\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

For example, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2} because they satisfy the equation when their values are substituted.

When x = -3, (-3)2 + 5 × -(3) + 6 = 0

When x = -2, (-2)2 + 5 × -(2) + 6 = 0

Without solving the equation 3x2 – 2x – 1 = 0, find whether x = 1 is a root or solution of this equation.

Solution:

Substituting x = 1 in the given equation 3x2 – 2x – 1 = 0, we get
3(1)2 – 2 (1) – 1 = 0
=> 3 – 2 – 1 = 0
=> 3 – 3 = 0
Therefore, x = 1 is a solution of the given equation 3x2 – 2x – 1 = 0

## How to Find the Roots of a Quadratic Equation

We can solve the quadratic equation to find its roots in different ways.

As we know α and β are the two roots of the quadratic equation, whose values can be determined using the quadratic formula:

(α, β) = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Let us verify if, indeed, the roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}

In the equation x2 + 5x + 6 = 0, a = 1, b = 5, c = 6

(α, β) = ${\dfrac{-5\pm \sqrt{\left( 5\right) ^{2}-4\times 1\times 6}}{-2\times 1}}$

= ${\dfrac{-5\pm \sqrt{25-24}}{2}}$

= ${\dfrac{-5+\sqrt{1}}{2}}$

= ${\dfrac{-5+1}{2}}$ and ${\dfrac{-5-1}{2}}$

= {-3, -2}

Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}

### By Factoring

Here we will factor the middle term ‘5x’ of the same equation x2 + 5x + 6 = 0

x2 + 5x + 6 = 0

=> x2 + 3x +2x + 6 = 0

=> x(x + 3) + 2(x + 3) = 0

=> (x + 3)(x + 2)=  0

=> x + 3 = 0 or x + 2 = 0

=> x = {-3, -2}

Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}

### By Completing the Square

Step 1: Isolating the x2 and x terms to one side of the equation

y = x2 + 5x + 6

=> y – 6 = x2 + 5x

Step 2: Add ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation

Here, b = ${\left( \dfrac{5}{2}\right) ^{2}}$ = 25/4

=> ${y-6+\dfrac{25}{4}=x^{2}+5x+\dfrac{25}{4}}$

Step 3: Factor the right side of the equation into a perfect square

=> ${y+\dfrac{1}{4}=\left( x+\dfrac{5}{2}\right) ^{2}}$

Step 4: Rewriting the equation in terms of y

=> ${y=\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}}$

When the quadratic function f(x) = y = 0

${\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}=0}$

=> ${\left( x+\dfrac{5}{2}\right) ^{2}=\dfrac{1}{4}}$

=> ${x+\dfrac{5}{2}=\pm \sqrt{\dfrac{1}{4}}}$

=> ${x=\left\{ +\dfrac{1}{2}-\dfrac{5}{2},-\dfrac{1}{2}-\dfrac{5}{2}\right\}}$

=> x= {-3, -2}

Thus, the two roots of the quadratic equation x2 + 5x + 6 = 0 are {-3, -2}

### By Graphing

Graphing the quadratic equation manually using a graphing calculator and then identifying the x-intercepts will give the roots of the quadratic equation.

The graph shows the two x-intercepts are (-2, 0) and (-3, 0). Thus the two roots of the quadratic equation are (-3, -2)

## Nature of Roots of the Quadratic Equation

The nature of the roots of the quadratic equation depends on the value of the discriminant as follows:

• If b2 – 4ac > 0, the quadratic equation has 2 real solutions
• If b2 – 4ac < 0, the quadratic equation has no real solutions but two different complex or imaginary roots
• If b2 – 4ac = 0, the quadratic equation has one real root

Find the value(s) of g if the quadratic equation 3y2 + gy + 2 = 0 has equal roots.

Solution:

As we know, thequadratic equation ax2 + bx + c = 0 has equal roots if its discriminant b2 – 4ac = 0
Here, a = 3, b = g, and c = 2
Putting the value in the discriminant, we get
b2 – 4ac = 0
=> (g)2 – 4 × 3 × 2 = 0
=> g2– 24 = 0
=> g2 = 24
=>g = ${\pm 2\sqrt{6}}$
Thus, when the given quadratic equation has equal roots, g = ${2\sqrt{6}}$ or g = ${-2\sqrt{6}}$

## Sum of the Roots

The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.

α + β = ${-\dfrac{b}{a}}$

Derivation

α + β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}+\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

= ${\dfrac{-b+\sqrt{b^{2}-4ac}-b-\sqrt{b^{2}-4ac}}{2a}}$

= ${-\dfrac{2b}{2a}}$ = ${-\dfrac{b}{a}}$

## Product of the Roots

The product of the roots of a quadratic equation is equal to the constant term divided by the leading coefficient.

α.β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}×\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

= ${\dfrac{b^{2}+b\sqrt{b^{2}-14ac}-b\sqrt{b^{2}-4ac}-\left( b^{2}-4ac\right) }{4a^{2}}}$

= ${\dfrac{b^{2}-b^{2}+4ac}{4a^{2}}}$

= ${\dfrac{4ac}{4a^{2}}}$ = ${\dfrac{c}{a}}$

Find the sum and the product of the roots of the quadratic equation a2 – 5a – 14 = 0

Solution:

Comparing the given quadratic equation with the standard form ax2 + bx + c = 0, here a = 1, b = -5, c = -14
As we know,
α + β = ${-\dfrac{b}{a}}$
= 1
Thus the sum of the roots of the quadratic equation a2 – 5a – 14 = 0 is 1
α.β = ${\dfrac{c}{a}}$
= -14
Thus the product of the roots of the quadratic equation a2 – 5a – 14 = 0 is -14

Write the quadratic equation with the given roots {7, -2}

Solution:

The quadratic equation can be obtained by multiplying the factors formed by the roots {7, -2}
The factors are (x – 3)(x + 2)
Thus, the quadratic equation is (x – 3)(x + 2) = 0
=> x2 + 2x – 3x – 6 = 0
=> x2 – x – 6 = 0