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.
Let \(x\mathbin{\spadesuit}y = x^2/y\) for all \(x\) and \(y\) such that \(y\neq 0\). Find all values of \(a\) such that \(a \mathbin{\spadesuit} (a + 1) = 9\). List the values you find in increasing order, separated by commas.
Alright, so in this case, we have \(x = a\) and \(y = a +1 \). Putting into the form of \(x\mathbin{\spadesuit}y = x^2/y\), we have \(a \mathbin{\spadesuit} (a + 1) = \frac{a^2}{a+1} = 9\).
Multiplying both sides by a + 1 and distributing the 9 in, we have \(a^2 = 9a + 9\). Now we can write the quadratic \(a^2 - 9a - 9 = 0\).
Unforunately, we can't factor this directly, so we have to use the quadratic formula to find that
\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)
This means our final answer is \(a=\frac{3\sqrt{13}+9}{2}, \frac{-3\sqrt{13}+9}{2}\)
Thanks!