Question

`x^2-2ax-(4b^2-a^2)=0`

Answer 1

The given equation is `x^2-2ax-(4b^2-a^2)=0` 

Comparing it with `Ax^2+Bx+C=0` we get 

`A=1,B=-2a and C=-(4b^2-a^2)` 

∴ Discriminant, 

`B^2-4AC=(-2a)^2-4xx1xx[-(4b^2-a^2)]=4a^2+16b^2-4a^2=16b^2>0` 

So, the given equation has real roots
Now, `sqrtD=sqrt16b^2=4b`  

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

`β=(-B-sqrt(D))/(2A)=(-(-2a)-4b)/(2xx1)=2(a-2b)/2=a-2b` 

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