3
$\begingroup$

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

$\endgroup$
0

4 Answers 4

Reset to default
4
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Okay. Got it. Is there a simple reminder rule why that is the case. $\endgroup$
    – Cetshwayo
    May 21, 2013 at 9:59
  • $\begingroup$ What do you mean? $\endgroup$
    – JSCB
    May 21, 2013 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, 2013 at 11:48
  • $\begingroup$ You are right$.$ $\endgroup$
    – JSCB
    May 21, 2013 at 12:11
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Got it thanks. Appreciate your help. $\endgroup$
    – Cetshwayo
    May 21, 2013 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, 2013 at 21:41
1
$\begingroup$

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
  • 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, 2013 at 13:29
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .