Question

`x^2-2ax+(a^2-b^2)=0` 

Answer 1

Given: 

`x^2-2ax+(a^2-b^2)=0` 

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

`A=1, B=-2a and C=(a^2-b^2)`  

Discriminant D is given by: 

`D=B^2-4AC` 

=`(-2a)^2-4xx1xx(a^2-b^2)`  

=`4a^2-4a^2+4b^2` 

=`4b^2>0` 

Hence, the roots of the equation are real. 

Roots α and β are given by:  

`α=(-b+sqrt(D))/(2a)=(-(-2a)+sqrt4b^2)/(2xx1)=(2a+2b)/2=(2(a+b))/2=(a+b)`

`β=(-b-sqrt(D))/(2a)=(-(-2a)-sqrt4b^2)/(2xx1)=(2a-2b)/2=(2(a-b))/2=(a-b)` 

Hence, the roots of the equation are (a +b) and (a +b).