Question

x⁴-16x²=0
4y⁴+7y²-2=0

Answer 1

$x^4-16x^=0\x^2(x^2-16)=0\x^2(x-4)(x+4)=0\x=0;x-4=0;x=4 x+4=0; x=-4\\4y^4+7y^2-2=0\z=y^2\4z^2+7z-2=0\D=49-4*4*(2)=81\z_{1} =frac{-7+sqrt{81} }{2*4} =frac{-7+9}{8} =frac{1}{4} \z_{2} =frac{-7-sqrt{81} }{2*4} =frac{-7-9}{8} =-2\y^2=frac{1}{4}\y=frac{1}{2} \y=-frac{1}{2}$