Algebra

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