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

$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ [CBSE 2015]

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Solution

The given equation is $x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ .

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

A = 1, B = $- (2 b - 1)$ and C = $b^{2} - b - 20$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (2 b - 1)]}^{2} - 4 \times 1 \times (b^{2} - b - 20) = 4 b^{2} - 4 b + 1 - 4 b^{2} + 4 b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{81} = 9$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] + 9}{2 \times 1} = \frac{2 b + 8}{2} = b + 4 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] - 9}{2 \times 1} = \frac{2 b - 10}{2} = b - 5$

Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.

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