Question

`a^2b^2x^2-(4b^4-3a^4)x-12a^2b^2=0,a≠0  and b≠ 0` 

 

Answer 1

The given equation is `a^2b^2x^2-(4b^4-3a^4)x-12a^2b^2=0` 

Comparing it with `Ax^2+Bx+C=0` 

`A=a^2b^2,B=-(4b^2-3a^4) and C=-12a^2b^2` 

∴ Discriminant, 

`B^2-4AC=[-(4b^4-3a^4)]^2-4xxa^2b^2xx(-12a^2b^2)=16b^8-24a^4b^4+9a^8+48a^4b^4` 

=`16b^8+24a^4b^4+9a^8=(4b^4+3a^4)^2>0` 

So, the given equation has real roots
Now, sqrtD=`sqrt(4b^4+3a^4)^2=4b^2+3a^4` 

∴` α =(-B+sqrt(D))/(2A)=(-[-(4b^4-3a^4)]+(4b^4+3a^4))/(2xxa^2b^2)=(8b^4)/(2a^2b^2)=(4b^2)/(a^2)` 

β=(-B-sqrt(D))/(2A)=(-[-(4b^4-3a^4)]+(4b^4+3a^4))/(2xxa^2b^2)=(-6b^4)/(2a^2b^2)=-(3a^2)/(a^2)` 

Hence, `(4b^2)/a^2`  and `(-3a^2)/(b^2)` are the roots of the given equation.