Question

`4x^2-4a^2x+(a^4-b^4)=0` 

 

Answer 1

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

Comparing it with `Ax^2+Bx+C=0` 

`A=4,B=-4a^2 and C=a^4-b^4` 

∴ Discriminant,` B^2-4AC=(-4a^2)^2-4xx4xx(a^2-b^4)=16a^4-16a^4=16b^4=16b^4>0` 

So, the given equation has real roots
Now,  `sqrtD=sqrt16b^4=4b^2` 

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

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

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