Question

`x^2-(2b-1)x+(b^2-b-20)=0` 

Answer 1

We write, -(2b-1)x=-(b-5)x-(b+4)x as 

x^2xx(b^2-b-20)=(b^2-b-20)x^2=[-(b-5)x]xx[-(b+4)x] 

∴ `x^2-(2b-1)x+(b^2-b-20)=0` 

⇒ `x^2-(b-5)x-(b+4)x+(b+4)=0` 

⇒`x[x-(b-5)]-(b+4)[x-(b-5)]=0` 

⇒` [x-b-5]x- [b+4]=0` 

⇒`x-(b-5)=0  or  x-(b+4)=0` 

⇒ `x=b-5  or x=b+4` 

Hence, b - 5 and b + 4 are the roots of the given equation.