Question

question 11
1 pts
solve for x. you may use your calculator or desmos.
$x + 2 = \sqrt{2x + 7}$

question 12
1 pts
solve for x. check for extraneous solutions.
$\sqrt{2x + 3} + 2 = \sqrt{6x + 7}$

Explanation:

Question 11

Step1: Square both sides

$$(x+2)^2 = (\sqrt{2x+7})^2$$
$$x^2 + 4x + 4 = 2x + 7$$

Step2: Rearrange to quadratic form

$$x^2 + 4x + 4 - 2x - 7 = 0$$
$$x^2 + 2x - 3 = 0$$

Step3: Factor the quadratic

$$(x+3)(x-1) = 0$$

Step4: Solve for x and check solutions

Test $x=-3$: $-3+2 = \sqrt{2(-3)+7} \implies -1 = \sqrt{1}$, which is false (extraneous).
Test $x=1$: $1+2 = \sqrt{2(1)+7} \implies 3 = \sqrt{9}$, which is true.

Step1: Isolate one square root

$$\sqrt{2x+3} = \sqrt{6x+7} - 2$$

Step2: Square both sides

$$(\sqrt{2x+3})^2 = (\sqrt{6x+7} - 2)^2$$
$$2x+3 = 6x+7 - 4\sqrt{6x+7} + 4$$

Step3: Simplify and isolate remaining root

$$2x+3 = 6x+11 - 4\sqrt{6x+7}$$
$$4\sqrt{6x+7} = 4x + 8$$
$$\sqrt{6x+7} = x + 2$$

Step4: Square both sides again

$$6x+7 = (x+2)^2$$
$$6x+7 = x^2 + 4x + 4$$

Step5: Rearrange to quadratic form

$$x^2 - 2x - 3 = 0$$

Step6: Factor and solve, then check

$$(x-3)(x+1) = 0$$
Test $x=-1$: $\sqrt{2(-1)+3}+2 = \sqrt{6(-1)+7} \implies \sqrt{1}+2 = \sqrt{1} \implies 3=1$, which is false (extraneous).
Test $x=3$: $\sqrt{2(3)+3}+2 = \sqrt{6(3)+7} \implies \sqrt{9}+2 = \sqrt{25} \implies 3+2=5$, which is true.

Answer:

$x=1$

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Question 12