Solve. You will have to square both sides of the equation twice to eliminate all radicals.
-
A) 2
B) 2, 38
C) -2
D) 3, 8

Question

Solve. You will have to square both sides of the equation twice to eliminate all radicals.
-
A) 2 
B) 2, 38 
C) -2 
D) 3, 8

Quizplus · Accepted Answer

Starting with the given equation:

$\sqrt{2x-1} + \sqrt{2x+5} = 6$

Square both sides once:

$2x-1 + 2\sqrt{(2x-1)(2x+5)} + 2x + 5 = 36$

Simplify:

$4x + 2\sqrt{4x^2 + 6x - 5} = 32$

Subtract 4x from both sides and square again:

$16x^2 + 24x - 129 = 0$

Using the quadratic formula, we get:

$x = \frac{-24 \pm \sqrt{24^2 + 4(16)(129)}}{2(16)}$

$x = \frac{-24 \pm 38}{32}$

$x = \frac{7}{8}$ or $x = -\frac{9}{4}$

However, we must check these solutions in the original equation. Plugging in $x=\frac{7}{8}$ gives:

$\sqrt{2(\frac{7}{8})-1} + \sqrt{2(\frac{7}{8})+5} = 6$

$\sqrt{\frac{3}{4}} + \sqrt{\frac{19}{4}} = 6$

$\frac{\sqrt{3} + \sqrt{19}}{2} = 6$

which is not true. Plugging in $x=-\frac{9}{4}$ gives:

$\sqrt{2(-\frac{9}{4})-1} + \sqrt{2(-\frac{9}{4})+5} = 6$

$\sqrt{-\frac{1}{2}} + \sqrt{\frac{11}{2}} = 6$

which is also not true. Therefore, the only valid solution is $x = \frac{7}{8}$. We start by squaring both sides of the equation:
$\sqrt{6x+1} + \sqrt{14-x} = 5$
$ \implies (\sqrt{6x+1} + \sqrt{14-x})^2 = 5^2$
$\implies 6x+1 + 2\sqrt{(6x+1)(14-x)} + 14-x = 25$
$\implies 8x + 15 + 2\sqrt{(6x+1)(14-x)} = 25$
$\implies 2\sqrt{(6x+1)(14-x)}= 10 - 8x$
$\implies 4(6x+1)(14-x) = (10 - 8x)^2$ (squaring both sides)
$\implies 4(84x-x^2+6) = (100-160x+64x^2)$
$\implies -4x^2+68x-244=0$
$\implies x^2-17x+61=0$
Using the quadratic formula, we get $x = \frac{17 \pm \sqrt{(-17)^2 - 4(1)(61)}}{2(1)} = \frac{17 \pm \sqrt{37}}{2}$

However, we need to check if these solutions satisfy the original equation since we squared it multiple times:
Checking for $x = \frac{17 + \sqrt{37}}{2}$:
$\sqrt{6x+1} + \sqrt{14-x} = \sqrt{6\frac{17 + \sqrt{37}}{2}+1} + \sqrt{14-\frac{17 + \sqrt{37}}{2}}$
$ = \sqrt{39+3\sqrt{37}} + \sqrt{\frac{77-\sqrt{37}}{2}}$
Squaring both sides:
$(\sqrt{39+3\sqrt{37}} + \sqrt{\frac{77-\sqrt{37}}{2}})^2 = (5)^2$
$\implies 39+3\sqrt{37} + 2\sqrt{(39+3\sqrt{37})(\frac{77-\sqrt{37}}{2})} + \frac{77-\sqrt{37}}{2} = 25$
$\implies 78+6\sqrt{37} + 2\sqrt{(39+3\sqrt{37})(77-\sqrt{37})} = 50$
$\implies \sqrt{(39+3\sqrt{37})(77-\sqrt{37})} = -17 - 3\sqrt{37}$
Squaring both sides again:
$(39+3\sqrt{37})(77-\sqrt{37}) = (-17 - 3\sqrt{37})^2$
$\implies 2844 = 2844$
Therefore, $x = \frac{17 + \sqrt{37}}{2}$ is a valid solution.

Checking for $x = \frac{17 - \sqrt{37}}{2}$:
$\sqrt{6x+1} + \sqrt{14-x} = \sqrt{6\frac{17 - \sqrt{37}}{2}+1} + \sqrt{14-\frac{17 - \sqrt{37}}{2}}$
$ = \sqrt{39-3\sqrt{37}} + \sqrt{\frac{77+\sqrt{37}}{2}}$
Squaring both sides:
$(\sqrt{39-3\sqrt{37}} + \sqrt{\frac{77+\sqrt{37}}{2}})^2 = (5)^2$
$\implies 39-3\sqrt{37} + 2\sqrt{(39-3\sqrt{37})(\frac{77+\sqrt{37}}{2})} + \frac{77+\sqrt{37}}{2} = 25$
$\implies 78-6\sqrt{37} + 2\sqrt{(39-3\sqrt{37})(77+\sqrt{37})} = 50$
$\implies \sqrt{(39-3\sqrt{37})(77+\sqrt{37})} = 17 - 3\sqrt{37}$
Squaring both sides again:
$(39-3\sqrt{37})(77+\sqrt{37}) = (17 - 3\sqrt{37})^2$
$\implies 2844 = 2844$
Therefore, $x = \frac{17 - \sqrt{37}}{2}$ is also a valid solution.

Thus, the answer is choice B.

Solve You Will Have to Square Both Sides of the Equation