Question
A positive real number is 4 less than another. When 8 times the larger is added to the square of the smaller, the result is 96.
A positive real number is 4 less than another. When 8 times the larger is added to the square of the smaller, the result is 96. Find the numbers.
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Koala Tutor
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Let the positive real numbers be $x$ and $y$. We form the system of equations \begin{align*}
y&=x-4,\\
8y&=x^2+96.
\end{align*} Substituting $x-4$ for $y$ in the second equation gives $8(x-4)=x^2+96 \Rightarrow x^2-8x+32=0.$ Factoring, we have $(x-4)(x-8)=0$. Since $x$ is positive, we have $x=8$. The smaller number is 4 less than the larger, so the smaller number is $8-4=\boxed{4}$.
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