Question

solve the radical equation \\(\sqrt{3x - 11} + 5 = x + 2\\). (1 point)\
\\(\bigcirc\\) \\(x = 4\\) or \\(x = -5\\)\
\\(\bigcirc\\) \\(x = -5\\)\
\\(\bigcirc\\) \\(x = 4\\) or \\(x = 5\\)\
\\(\bigcirc\\) \\(x = 4\\)

Explanation:

Step1: Isolate the radical term

Subtract 5 from both sides:
$\sqrt{3x - 11} = x + 2 - 5$
$\sqrt{3x - 11} = x - 3$

Step2: Square both sides

Eliminate the square root:
$(\sqrt{3x - 11})^2 = (x - 3)^2$
$3x - 11 = x^2 - 6x + 9$

Step3: Rearrange to quadratic form

Move all terms to one side:
$x^2 - 6x + 9 - 3x + 11 = 0$
$x^2 - 9x + 20 = 0$

Step4: Factor the quadratic

Find two factors of 20 that sum to -9:
$(x - 4)(x - 5) = 0$
Solutions from factoring: $x=4$ or $x=5$

Step5: Verify solutions

Check $x=4$ in original equation:
$\sqrt{3(4)-11} +5 = 4+2$
$\sqrt{1} +5 = 6$
$1+5=6$, which is true.

Check $x=5$ in original equation:
$\sqrt{3(5)-11} +5 = 5+2$
$\sqrt{4} +5 = 7$
$2+5=7$, which is true.

Answer:

$x=4$ or $x=5$ (corresponding to the option: $x = 4$ or $x = 5$)