Asked by Bob

If x≠0, x/2=y^2 and x/4=4y, then what is the x
Answered by Reiny
if x/2 = y^2 ----> x = 2y^2
if x/4 = 4y ----> x = 16y

then 2y^2 = 16y
2y^2 - 16y =0
2y(y - 8 ) = 0
y = 0 or y = 8
then in x = 2y^2
x = 0 or x = 128
but x ≠ 0,
so x = 128
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