Question

`4x^2-4bx-(a^2-b^2)=0` 

 

Answer 1

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

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

A=4,B=4b and C=`-(a^2-b^2)` 

`A=4,B=4b and C=-(a^2-b^2)` 

∴ Discriminant, 

`D=B^2-4AC=(4b)^2-4xx4xx[-(a^2-b^2)]=16b^2+16a^2-16b^2=16a^2>0` 

So, the given equation has real roots
Now, `sqrtD=sqrt16a^2=4a` 

∴` α=((-B+sqrt(D))/(2A))=(-4b+4a)/(2xx4)=(4(a-b))/8=(a-b)/2` 

`β=((-B-sqrt(D))/(2A))=(-4b-4a)/(2xx4)=(4(a+b))/8=(a+b)/2`  

Hence, `1/2(a-b)` and `-1/2(a+b)` are the roots of the given equation.