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 
 A: When you square the LHS, it should be $9(x+13)$ not $9+x+13$.
A: When you square both sides you should get $9(x+13) = (x+9)^2$ which rearranges to
\[x^2+9x-36 = 0 \ ,\]
which has the solutions
\[x_{1,2} = -\frac 92 \pm \sqrt{\frac{81}{4}+36} = \frac{-9\pm 15}{2} \: ,\]
i.e. $\begin{cases}x_1 = 3 \\ x_2 = -12\end{cases}$. By putting these into the original equation $3\sqrt{x+13} = x+9$ you realize that $x_2 = -12$ is not a solution, but $x_1 = 3$ is.
A: 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.
A: 3 √x+13 = x + 9
Squaring both sides, we get
9 (x + 13) = x^2 + 81 + 18x
9x + 117 = x^2 + 81 + 18x
x^2 + 18x - 9x + 81 - 117 = 0
x^2 + 9x - 36 = 0
x^2 + 12x - 3x - 36 = 0
x(x + 12) - 3(x + 12) = 0
(x + 12)(x - 3) = 0
Therefore, x = -12 or 3
However, when the problem equation does not get satisfied when x is substituted by -12, but it gets satisfied when x is substituted by 3.
Therefore, 3 is the required solution or root of the equation.
Hope that you understood the answer. 
