Positive real numbers $r,s$ satisfy the equations $r^2 + s^2 = 1$ and $r^4 + s^4= \frac{7}{8}$.  Find $rs$.
We have $2r^2s^2 = (r^4 + 2r^2s^2 + s^4) - (r^4 + s^4) = (r^2 + s^2)^2 - (r^4 + s^4) = (1)^2 - \frac{7}{8} = \frac{1}{8}$, so $r^2s^2 = \frac{1}{16}$. This means that $rs = \boxed{\frac{1}{4}}$.
\frac{1}{4}
Level 4
Algebra
Problem: Positive real numbers $r,s$ satisfy the equations $r^2 + s^2 = 1$ and $r^4 + s^4= \frac{7}{8}$.  Find $rs$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( rs \), we start with the given equations \( r^2 + s^2 = 1 \) and \( r^4 + s^4 = \frac{7}{8} \). 

First, we can express \( r^4 + s^4 \) in terms of \( r^2 \) and \( s^2 \). We know that:

\[
r^4 + s^4 = (r^2 + s^2)^2 - 2r^2s^2
\]

Substituting \( r^2 + s^2 = 1 \) into the equation gives:

\[
r^4 + s^4 = 1^2 - 2r^2s^2 = 1 - 2r^2s^2
\]

Setting this equal to the second equation, we have:

\[
1 - 2r^2s^2 = \frac{7}{8}
\]

To isolate \( r^2s^2 \), we rearrange the equation:

\[
1 - \frac{7}{8} = 2r^2s^2
\]

This simplifies to:

\[
\frac{1}{8} = 2r^2s^2
\]

Dividing both sides by 2 yields:

\[
r^2s^2 = \frac{1}{16}
\]

Next, we can express \( rs \) in terms of \( r^2s^2 \):

\[
(rs)^2 = r^2s^2
\]

Taking the square root of both sides gives:

\[
rs = \sqrt{r^2s^2} = \sqrt{\frac{1}{16}} = \frac{1}{4}
\]

Thus, the value of \( rs \) is confirmed to be \( \frac{1}{4} \).<|END_OF_LATENT|> The final answer is $\frac{1}{4}$.