Question

y varies directly as the square of x. When x = 1, y = 8. Find x when y = 128.
x = ±4

x = ±6

x = ±8

x = ±16
what does this mean

Answer 1

this means \[y=k \times x^2\]

Answer 2

so\[8=k \times 1^2\]

Answer 3

can you work out what the value of k is

Answer 4

\[128=8\times x^2\]
\[128\div8=x^2\]
\[16=x^2\]
x=4

Answer 5

actually its \[\pm4\] as -4 squared = 16