Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ [CBSE 2015]

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Solution

The given equation is $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 6 and C = $- (a^{2} + 2 a - 8)$

∴ Discriminant, D = $B^{2} - 4 A C = 6^{2} - 4 \times 1 \times [- (a^{2} + 2 a - 8)] = 36 + 4 a^{2} + 8 a - 32 = 4 a^{2} + 8 a + 4 = 4 (a^{2} + 2 a + 1) = 4 {(a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 {(a + 1)}^{2}} = 2 (a + 1)$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 6 + 2 (a + 1)}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 6 - 2 (a + 1)}{2 \times 1} = \frac{- 2 a - 8}{2} = - a - 4 = - (a + 4)$

Hence, $(a - 2)$ and $- (a + 4)$ are the roots of the given equation.

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