Question

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

x = ±4

x = ±6

x = ±8

x = ±12

Answer 1

y varies directly as the square of x so

y = kx^2

when x = 4 then y = 8

y = kx^2
8 = k(4)^2
8 = 16k
8/16 = k
1/2 = k

what is x when y = 32

y = kx^2
32 = (1/2)x^2
(32 x 2) = x^2
\(\small \sqrt{32 \times 2}\) = x
does that help?