Question

`36x^2-12ax+(a^2-b^2)=0` 

Answer 1

The given equation is `36x^2-12ax(a^2-b^2)=0` 

Comparing it with `Ax^2+Bx+C=0` 

`A=36,B=-12a  and C=a^2-b^2` 

∴ Discriminant, 

`D=B^2-4AC=(-12a)^2-4xx36xx(a^2-b^2)=144a^2-144a^2+144b^2=144b^2>0` 

So, the given equation has real roots
Now,  `sqrtD=sqrt144b^2=12b` 

∴ `α =(-B+sqrtD)/(2A)=(-(-12a)+12b)/(2xx36)=(12(a+b))/72=(a+b)/0`

β=`(-B-sqrtD)/(2A)=(-(-12a)-12b)/(2xx36)=(12(a-b))/72=(a-b)/0` 

Hence, `(a+b)/6` and `(a-b)/6` are the roots of the given equation.