Question

solve the radical equation, \\(\sqrt{q - 2} + 3 = \sqrt{4q + 1}\\). select one: \\(\bigcirc\\) a. no solution. \\(\bigcirc\\) b. \\(q = 2, 6\\) \\(\bigcirc\\) c. \\(q = 2\\) \\(\bigcirc\\) d. \\(q = 6\\)

Explanation:

Step1: Isolate one radical term

$\sqrt{q-2} = \sqrt{4q+1} - 3$

Step2: Square both sides

$(\sqrt{q-2})^2 = (\sqrt{4q+1} - 3)^2$
$q - 2 = (4q + 1) - 6\sqrt{4q+1} + 9$

Step3: Simplify the equation

$q - 2 = 4q + 10 - 6\sqrt{4q+1}$
$-3q - 12 = -6\sqrt{4q+1}$
Divide both sides by $-3$:
$q + 4 = 2\sqrt{4q+1}$

Step4: Square both sides again

$(q + 4)^2 = (2\sqrt{4q+1})^2$
$q^2 + 8q + 16 = 4(4q + 1)$
$q^2 + 8q + 16 = 16q + 4$

Step5: Rearrange to quadratic form

$q^2 - 8q + 12 = 0$
Factor:
$(q - 2)(q - 6) = 0$
Potential solutions: $q=2, q=6$

Step6: Verify solutions

For $q=2$:
Left side: $\sqrt{2-2}+3=0+3=3$
Right side: $\sqrt{4*2+1}=\sqrt{9}=3$ (valid)
For $q=6$:
Left side: $\sqrt{6-2}+3=2+3=5$
Right side: $\sqrt{4*6+1}=\sqrt{25}=5$ (valid)

Answer:

B. $q = 2, 6$