2
$\begingroup$

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

$\endgroup$
3
$\begingroup$

When you square the LHS, it should be $9(x+13)$ not $9+x+13$.

$\endgroup$
  • $\begingroup$ Okay. Got it. Is there a simple reminder rule why that is the case. $\endgroup$ – Cetshwayo May 21 '13 at 9:59
  • $\begingroup$ What do you mean? $\endgroup$ – A. Chu May 21 '13 at 10:06
  • 1
    $\begingroup$ $3 \sqrt{x+13}$ is a product of the form $ab$, therefore its square is of the form $a^2 b^2$ $\endgroup$ – mau May 21 '13 at 11:48
  • $\begingroup$ You are right$.$ $\endgroup$ – A. Chu May 21 '13 at 12:11
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Got it thanks. Appreciate your help. $\endgroup$ – Cetshwayo May 21 '13 at 10:04
  • $\begingroup$ You can avoid substituting in the original equation by just observing that a solution $x$ must satisfy $x>-9$ and this is the only additional condition. Back substitution should be the last resort. $\endgroup$ – egreg May 21 '13 at 21:41
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I suppose that $9(x+13) = \dfrac{x^2+18x+81}{9}$ is a typo and you wanted to write $9(x+13) = x^2+18x+81$. $\endgroup$ – Martin Sleziak Nov 28 '13 at 13:29
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.