Question

`x^2+6x-(a^2+2a-8)=0` 

Answer 1

The given equation is `x^2+6x-(a^2+2a-8)=0` 

Comparing it with `Ax^2+Bx+C=0` 

`A=1,B=6 and C=-(a^2+2a-8)` 

∴ Discriminant, D= 

`B^2-4AC=6^2-4xx1xx[-(a^2+2a-8)]=36+4a^2+8a-32=4a^2+8a-32=4a^2+8a+4` 

`=4(a^2+2a+1)=4(a+1)^2>0` 

So, the given equation has real roots 
Now, `sqrtD=sqrt(4(a+1)^2)=2(a+1)`  

∴`α=(-B+sqrt(D))/(2A)=(-6+2(a+1))/(2xx1)=(2a-4)/2=a-2` 

`β=(-B-sqrt(D))/(2A)=(-6-2(a+1))/(2xx1)=(2a-8)/2=a-4=-(a+4)` 

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