Examine the Roots of a Quadratic Equation

Examining the roots of a quadratic equation means to see the type of its roots i.e., whether they are real or imaginary, rational or irrational, equal or unequal.

The nature of the roots of a quadratic equation depends entirely on the value of its discriminant b $^{2}$ - 4ac.

In a quadratic equation ax $^{2}$ + bx + c = 0, a ≠ 0 the coefficients a, b and c are real. We know, the roots (solution) of the equation ax $^{2}$ + bx + c = 0 are given by x = $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ .

1. If b $^{2}$ - 4ac = 0 then the roots will be x = $\frac{-b ± 0}{2a}$ = $\frac{-b - 0}{2a}$ , $\frac{-b + 0}{2a}$ = $\frac{-b}{2a}$ , $\frac{-b}{2a}$ .

Clearly, $\frac{-b}{2a}$ is a real number because b and a are real.

Thus, the roots of the equation ax $^{2}$ + bx + c = 0 are real and equal if b $^{2}$ – 4ac = 0.

2. If b $^{2}$ - 4ac > 0 then $\sqrt{b^{2} - 4ac}$ will be real and non-zero. As a result, the roots of the equation ax $^{2}$ + bx + c = 0 will be real and unequal (distinct) if b $^{2}$ - 4ac > 0.

3. If b $^{2}$ - 4ac < 0, then $\sqrt{b^{2} - 4ac}$ will not be real because $(\sqrt{b^{2} - 4ac})^{2}$ = b $^{2}$ - 4ac < 0 and square of a real number always positive.

Thus, the roots of the equation ax $^{2}$ + bx + c = 0 are not real if b $^{2}$ - 4ac < 0.

As the value of b $^{2}$ - 4ac determines the nature of roots (solution), b $^{2}$ - 4ac is called the discriminant of the quadratic equation.

For the quadratic equation ax $^{2}$ + bx + c =0, a ≠ 0; the expression b $^{2}$ - 4ac is called discriminant and is, in general, denoted by the letter ‘D’.

Thus, discriminant D = b $^{2}$ - 4ac

Note:

Discriminant of ax $^{2}$ + bx + c = 0	Nature of roots of ax $^{2}$ + bx + c = 0	Value of the roots of ax $^{2}$ + bx + c = 0
b $^{2}$ - 4ac = 0	Real and equal	- $\frac{b}{2a}$ , - $\frac{b}{2a}$
b $^{2}$ - 4ac > 0	Real and unequal	$\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
b $^{2}$ - 4ac < 0	Not real	No real value

When a quadratic equation has two real and equal roots we say that the equation has only one real solution.

Solved examples to examine the nature of roots of a quadratic equation:

1. Prove that the equation 3x $^{2}$ + 4x + 6 = 0 has no real roots.

Solution:

Here, a = 3, b = 4, c = 6.

So, the discriminant = b $^{2}$ - 4ac

= 4 $^{2}$ - 4 ∙ 3 ∙ 6 = 36 - 72 = -56 < 0.

Therefore, the roots of the given equation are not real.

2. Find the value of ‘p’, if the roots of the following quadratic equation are equal (p - 3)x $^{2}$ + 6x + 9 = 0.

Solution:

For the equation (p - 3)x $^{2}$ + 6x + 9 = 0;

a = p - 3, b = 6 and c = 9.

Since, the roots are equal

Therefore, b $^{2}$ - 4ac = 0

⟹ (6) $^{2}$ - 4(p - 3) × 9 = 0

⟹ 36 - 36p + 108 = 0

⟹ 144 - 36p = 0

⟹ -36p = - 144

⟹ p = $\frac{-144}{-36}$

⟹ p = 4

Therefore, the value of p = 4.

3. Without solving the equation 6x $^{2}$ - 7x + 2 = 0, discuss the nature of its roots.

Solution:

Comparing 6x $^{2}$ - 7x + 2 = 0 with ax $^{2}$ + bx + c = 0 we have a = 6, b = -7, c = 2.

Therefore, discriminant = b $^{2}$ – 4ac = (-7) $^{2}$ - 4 ∙ 6 ∙ 2 = 49 - 48 = 1 > 0.

Therefore, the roots (solution) are real and unequal.

Note: Let a, b and c be rational numbers in the equation ax $^{2}$ + bx + c = 0 and its discriminant b $^{2}$ - 4ac > 0.

If b $^{2}$ - 4ac is a perfect square of a rational number then $\sqrt{b^{2} - 4ac}$ will be a rational number. So, the solutions x = $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ will be rational numbers. But if b $^{2}$ – 4ac is not a perfect square then $\sqrt{b^{2} - 4ac}$ will be an irrational numberand as a result the solutions x = $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ will be irrational numbers. In the above example we found that the discriminant b $^{2}$ – 4ac = 1 > 0 and 1 is a perfect square (1) $^{2}$ . Also 6, -7 and 2 are rational numbers. So, the roots of 6x $^{2}$ – 7x + 2 = 0 are rational and unequal numbers.