Question

`2x^2+ax-a^2=0`

Answer 1

The given equation is `2x^ax-a^2=0` 

Comparing it with `Ax^2+bx+c=0` 

`A=2,B=a and C=-a^2` 

∴ Discriminant, `D= B^2-4AC=a^2-4xx2xxa^2=a^2+8a^2=9a^2≥0` 

So, the given equation has real roots.
Now, `sqrtD=sqrt9a^2=3a` 

`α=(-B+sqrtD)/(2A)=(-a+3a)/(2xx2)=(2a)/4=4/2` 

`β=(-B+sqrtD)/(2A)=(-a+3a)/(2xx2)=(-4a)/4=-a` 

Hence, `a/2` and `-a` are the roots of the given equation.