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Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: x^{2}+6 x-\left(a^{2}+2 a-8\right)=0.
CBSE | Class 10 | Exercise 10C | Maths | Quadratic Equations | RS Aggarwal
Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: x^{2}+6 x-\left(a^{2}+2 a-8\right)=0.

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

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

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

\therefore Discriminant, D=

\begin{array}{l} B^{2}-4 A C=6^{2}-4 \times 1 \times\left[-\left(a^{2}+2 a-8\right)\right]=36+4 a^{2}+8 a-32=4 a^{2}+8 a-32=4 a^{2}+8 a+4 \\ =4\left(a^{2}+2 a+1\right)=4(a+1)^{2}>0 \end{array}

So, the given equation has real roots

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

\begin{array}{l} \therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-6+2(a+1)}{2 \times 1}=\frac{2 a-4}{2}=a-2 \\ \beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-6-2(a+1)}{2 \times 1}=\frac{-2 a-8}{2}=-a-4=-(a+4) \end{array}

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

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