Question

y varies directly as the square of x. when x=4, y=8. find x when y=32

Answer 1

y=kx^2, where k is a constant
when x=4,y=8
8=k4^2
8=16k
k=8/16=1/2
\[y=\frac{ 1 }{ 2 }x ^{2},when y=32,\]
\[x ^{2}=2y=2*32=64,x=\pm8\]