Question

P^2q^2x^2-(4q^4-3p^4)x-12p^2q^2=0 using quadratic formula

Answer 1

`p^2q^2x^2-(4q^4-3p^4)x-12p^2q^2=0\\\\a=p^2q^2,\ b=-(4q^4-3p^4);\ c=-12p^2q^2\\\\b^2-4ac=[-(4q^4-3p^4)]^2-4(p^2q^2)(-12p^2q^2)\\\\=(4q^4)^2-2(4q^4)(3p^4)+(3p^4)^2+48p^4q^4\\\\=16q^8-24q^4p^4+9p^8+48p^4q^4=16q^8+24p^4q^4+9p^8`

`x_1=(-b-√(b^2-4ac))/(2a);\ x_2=(-b+√(b^2-4ac))/(2a)\\\\x_1=(4q^4-3p^4-√(16q^8+24p^4q^4+9p^8))/(2p^2q^2)\\\\x_2=(4q^4-3p^4+√(16q^8+24p^4q^4+9p^8))/(2p^2q^2)\\\\\text{The domain}\\D:p\\eq0\ \wedge\ q\\eq0`