Question

`x^2+5x-(a^+a-6)=0` 

 

Answer 1

The given equation is `x^2+5x-(a^2+a-6)=0` 

Comparing it with `Ax^2+Bx+C=0` 

`A=1,B=5 and C=-(a^2+a-6)` 

∴ Discriminant, D = 

`B^2-4AC=5^2-4xx1xx[-(a^2+a-6)]=25+4a^2+4a-24=4a^2+4a^2+4a+1` 

=`(2a+1)^2>0`  

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

∴`α=(-B+sqrtD)/(2A)=(-5+2a+1)/(2xx1)=(2a-4)/2=a-2`

`β=(-B-sqrtD)/(2A)=(-5-2a+1)/(2xx1)=(-2a-6)/2=a-2=-a-3=-(a+3)` 

Hence, (a-2) and -(a+3) are the roots of the given equation.