Question

x⁴-16x²=0
4y⁴+7y²-2=0

Answer 1

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