Math Help - Regarding direct proportions

1. Regarding direct proportions

If y is directly proportional to x^2 and y=1/8 when x=1/2, what is the positive value of x when y=9/2?

a) 3/4
b) 3/2
c) 9/4
d) 3
e) 9

what i did:
y: 1/8
x^2: 1/4

1/8 * 36 = 9/2

therefore

1/4 * 36 = 9

but the answer is 3, not 9...

2. Hello juliak
Originally Posted by juliak
If y is directly proportional to x^2 and y=1/8 when x=1/2, what is the positive value of x when y=9/2?

a) 3/4
b) 3/2
c) 9/4
d) 3
e) 9

what i did:
y: 1/8
x^2: 1/4

1/8 * 36 = 9/2

therefore

1/4 * 36 = 9

but the answer is 3, not 9...
Translate the phrase " $y$ is directly proportional to $x^2$" in symbols as:
$y = kx^2$, where $k$ is a constant
Then plug in the two values that you've been given:

When $y = \tfrac18, x= \tfrac12$. So:
$\tfrac18=k(\tfrac12)^2$
$=\tfrac14k$
$\Rightarrow k = \tfrac12$
So the equation is:
$y = \tfrac12x^2$
When $y = \tfrac92$:
$\tfrac92=\tfrac12x^2$

$\Rightarrow x^2 = 9$

$\Rightarrow x = 3$

3. Originally Posted by juliak
If y is directly proportional to x^2 and y=1/8 when x=1/2, what is the positive value of x when y=9/2?

a) 3/4
b) 3/2
c) 9/4
d) 3
e) 9

what i did:
y: 1/8
x^2: 1/4

1/8 * 36 = 9/2

therefore

1/4 * 36 = 9

but the answer is 3, not 9...
If y is directly proportional to x² both values have to satisfy the equation:

$y = k \cdot x^2$

for a certain proportionality factor k.

You know

$\frac18 = k \cdot \left( \frac12 \right)^2$

Solve for k.

Then plug in $y = \frac92$ into the equation and solve for x.

4. Originally Posted by juliak
If y is directly proportional to x^2 and y=1/8 when x=1/2, what is the positive value of x when y=9/2?

a) 3/4
b) 3/2
c) 9/4
d) 3
e) 9

what i did:
y: 1/8
x^2: 1/4

1/8 * 36 = 9/2

therefore

1/4 * 36 = 9

but the answer is 3, not 9...
Because, just as "1/4" was x^2= (1/2)^2 so is this 9= x^2. You need to take the square root to get x.