Question

question 4 of 10, step 1 of 1 solve the following radical equation. (sqrt{6x + 21}+3 = x + 2) answer how to enter your answer (opens in new window) write your answer(s) beginning with the first answer box. if applicable, the second answer box may be left blank. keyboard shortcuts keypad (x=) ,

Explanation:

Step1: Isolate the radical term

Subtract 3 from both sides to get the square root alone.
$\sqrt{6x+21} = x + 2 - 3$
$\sqrt{6x+21} = x - 1$

Step2: Square both sides

Eliminate the square root by squaring each side.
$(\sqrt{6x+21})^2 = (x - 1)^2$
$6x + 21 = x^2 - 2x + 1$

Step3: Rearrange to quadratic form

Move all terms to one side to set equation to 0.
$x^2 - 2x + 1 - 6x - 21 = 0$
$x^2 - 8x - 20 = 0$

Step4: Factor the quadratic

Find two factors of -20 that sum to -8.
$(x - 10)(x + 2) = 0$

Step5: Solve for x

Set each factor equal to 0 and solve.
$x - 10 = 0 \implies x = 10$
$x + 2 = 0 \implies x = -2$

Step6: Verify solutions

Check each solution in the original equation.
For $x=10$:
$\sqrt{6(10)+21} + 3 = \sqrt{81} + 3 = 9 + 3 = 12$, and $10 + 2 = 12$. This is valid.

For $x=-2$:
$\sqrt{6(-2)+21} + 3 = \sqrt{9} + 3 = 3 + 3 = 6$, but $-2 + 2 = 0$. $6
eq 0$, so this is an extraneous solution.

Answer:

$x = 10$