Question

question 6
which of the following is the extraneous solution of the radical equation \\(\sqrt{2x - 3} + 3 = x\\) ?
a \\(x = 0\\)
b \\(x = -2\\)
c \\(x = 2\\)
d \\(x = 6\\)
question 7
last saved 10:47:43 pm
questions filter (11)

Explanation:

Step1: Isolate the radical

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

Step2: Square both sides

$(\sqrt{2x - 3})^2 = (x - 3)^2$ → $2x - 3 = x^2 - 6x + 9$

Step3: Rearrange into quadratic

$x^2 - 8x + 12 = 0$

Step4: Factor the quadratic

$(x - 2)(x - 6) = 0$ → $x = 2$ or $x = 6$

Step5: Check solutions

• For $x = 2$: $\sqrt{2(2) - 3} + 3 = \sqrt{1} + 3 = 4

eq 2$ (extraneous)

• For $x = 6$: $\sqrt{2(6) - 3} + 3 = \sqrt{9} + 3 = 6$ (valid)

Answer:

C. \( x = 2 \)