Solve each of the following quadratic equations:
Question:

Solve each of the following quadratic equations:

$x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$

 

Solution:

We write, $6 x=(a+4) x-(a-2) x$ as $x^{2} \times\left[-\left(a^{2}+2 a-8\right)\right]=-\left(a^{2}+2 a-8\right) x^{2}=(a+4) x \times[-(a-2) x]$

$\therefore x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$

$\Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0$

$\Rightarrow x[x+(a+4)]-(a-2)[x+(a+4)]=0$

$\Rightarrow[x+(a+4)][x-(a-2)]=0$

$\Rightarrow x+(a+4)=0$ or $x-(a-2)=0$

$\Rightarrow x=-(a+4)$ or $x=a-2$

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

 

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