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.
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! :)
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