<s>[INST] Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$ 
However, the answer is 3 
 [/INST] Given: $\boxed{3\sqrt{x+13} = x+9}$
Isolate the radical, x's cancel:
$\dfrac{3\sqrt{x+13}}{3} = \dfrac{x+9}{3}$
$\sqrt{x+13} = \dfrac{x+9}{3}$
Square both sides:
$\left(\sqrt{x+13}\right)^2 =\left(\dfrac{x+9}{3}\right)^2$
$x+13 = \left(\dfrac{x+9}{3}\right)\left(\dfrac{x+9}{3}\right)$
Move the 9 to the other side:
$9(x+13) = \dfrac{x^2+18x+81}{9}$
Set the equation equal to zero and group like terms:
$9x+117 = x^2+18x+81$
$0 = x^2+18x-9x+81-117$
Factor:
$0 = x^2+9x-36$
$0=(x+12)(x-3)$
$0=x+12\implies \boxed{x=-12}$
$0=x-3\implies\boxed{x=3}$
We can check to see if these are solutions:
$3\sqrt{-12+13} = -12+9$
$3(1)=-3$
$\boxed{3 \neq -3}$
Therefore, -12 is not a solution to the original equation.
$3\sqrt{3+13} = 3+9$
$3(4)=12$
$\boxed{12= 12}$
Therefore, 3 is the solution to the original equation.
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