Question

`x^2-(2b-1)x+(b^2-b-20)=0`

Answer 1

The given equation is `x^2-(2b-1)x+(b^2-b-20)=0` 

Comparing it with `Ax^2+Bx+C=0` 

`A=1,B=-(2b-1) and C=b^2-b-20` 

∴ Discriminant, 

`D=B^2-4AC=[-(2b-1)]^2-4xx1xx(b^2-b-20)=4b^2-4b+1-4b^2+4b+80=81>0` 

So, the given equation has real roots
Now, `sqrtD=sqrt18=9` 

∴α=`(-B+sqrt(D))/(2A)=(-[-(2b-1)]+9)/(2xx1)=(2b+8)/2=b+4`  

`β=(-B-sqrt(D))/(2A)=(-[-(2b-1)]-9)/(2xx1)=(2b-10)/2=b-5`  

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