Question

What is the solution for x in the given equation? (root)9x+7+ (root)2x=7 A. x = 18 and x = 2 B. x = 18 C. x = 2 D. x = 18 and x = -2

Answer 1

Answer:

C. x = 2

Step-by-step explanation:

$\sqrt{9x + 7} + \sqrt{2x} = 7$

Since you have square roots, you need to separate the square roots and square both sides.

$\sqrt{9x + 7} = 7 - \sqrt{2x}$

Now that one square root is on each side of the equal sign, we square both sides.

$(\sqrt{9x + 7})^2 = (7 - \sqrt{2x})^2$

$9x + 7 = 49 - 14\sqrt{2x} + 2x$

Now we isolate the square root and square both sides again.

$7x - 42 = -14\sqrt{2x}$

Every coefficient is a multiple of 7, so to work with smaller numbers, we divide both sides by 7.

$x - 6 = -2\sqrt{2x}$

Square both sides.

$(x - 6)^2 = (-2\sqrt{2x})^2$

$x^2 - 12x + 36 = 4(2x)$

$x^2 - 20x + 36 = 0$

We need to try to factor the left side.

-2 * (-18) = 36 & -2 + (-18) = -20, so we use -2 and -18.

$(x - 2)(x - 18) = 0$

$x = 2$   or   $x = 18$

Since solving this equation involved the method of squaring both sides, we much check for extraneous solutions by testing our two solutions in the original equation.

Test x = 2:

$\sqrt{9x + 7} + \sqrt{2x} = 7$

$\sqrt{9(2) + 7} + \sqrt{2(2)} = 7$

$\sqrt{25} + \sqrt{4} = 7$

$5 + 2 = 7$

$5 = 5$

We have a true equation, so x = 2 is a true solution of the original equation.

Now we test x = 18.

$\sqrt{9x + 7} + \sqrt{2x} = 7$

$\sqrt{9(18) + 7} + \sqrt{2(18)} = 7$

$\sqrt{162 + 7} + \sqrt{36} = 7$

$\sqrt{169} + 6 = 7$

$13 + 6 = 7$

$19 = 7$

Since 19 = 7 is a false equation, x = 18 is not a true solution of the original equation and is discarded as an extraneous solution.

Answer: C. x = 2