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Let x \mathbin{\spadesuit} y = \frac{x^2}{y} for all x and y such that y\neq 0. Find all values of $a$ such that $a \mathbin{\spadesuit} (a + 1) = 9$. Write your answer as a list separated by commas.

 
 Jun 11, 2024
 #1
avatar+759 
+1

We want to solve the equation for

\(a\mathbin{\spadesuit} (a + 1) = 9\)

 

Now, we plug in the function, and we get the equation \(9=\frac{a^2}{a+1}\)

 

Now, we simplfy solve the equation. We get

\(9a+9=a^2\)

\(a^2-9a-9=0 \)

 

Using the quadratic equation, we get

\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)

 

So our answer is

\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)

 

Thanks! :)

 Jun 11, 2024
 #2
avatar+74 
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If \(x \mathbin{\spadesuit} y = \frac{x^2}{y}\), then \(a \mathbin{\spadesuit} (a + 1) = \frac{a^2}{a+1}\). Therfore, \(\frac{a^2}{a+1}=9\). We can now use completing the square to solve for a.

 

\(\frac{a^2}{a+1}=9\)

\(a^2 = 9a+9\)

\(a^2-9a-9=0\)

\(a^2-9a+20.25 = 29.25\)

\((a-4.5)^2 = 29 \frac{1}{4}\)

\((a-\frac{9}{2})^2 = \frac{117}{4}\)

\(a-\frac{9}{2}=\frac{\pm\sqrt{117}}{2}\)

\(\mathbf{a = \frac{9\pm3\sqrt{13}}{2}}\)

 

So the two solutions are (9 + 3sqrt(13))/2, (9 - 3sqrt(13))/2

 Jun 11, 2024
edited by Maxematics  Jun 11, 2024

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