Question

for the following exercise, solve the radical equation. be sure to check all solutions to eliminate extraneous solutions.\\(\sqrt{2x + 3} - 1 = 0\\)\\(x = \\)\\(\\)\\(\\)question 24\\(\\)for the following exercise, solve the radical equation. be sure to check all solutions to eliminate extraneous solutions.\\(\sqrt{2x + 15} - \sqrt{x + 3} = 3\\)\\(x = \\)\\(\\)

Explanation:

Question 23

Step1: Isolate the radical term

$\sqrt{2x+3} = 1$

Step2: Square both sides

$(\sqrt{2x+3})^2 = 1^2$
$2x + 3 = 1$

Step3: Solve for $x$

$2x = 1 - 3$
$2x = -2$
$x = -1$

Step4: Verify the solution

Substitute $x=-1$ into original equation:
$\sqrt{2(-1)+3} - 1 = \sqrt{1} - 1 = 1 - 1 = 0$, which matches the right-hand side.

Question 24

Step1: Isolate one radical term

$\sqrt{2x+15} = 3 + \sqrt{x+3}$

Step2: Square both sides

$(\sqrt{2x+15})^2 = (3 + \sqrt{x+3})^2$
$2x + 15 = 9 + 6\sqrt{x+3} + x + 3$

Step3: Simplify and isolate remaining radical

$2x + 15 = x + 12 + 6\sqrt{x+3}$
$x + 3 = 6\sqrt{x+3}$

Step4: Square both sides again

$(x + 3)^2 = (6\sqrt{x+3})^2$
$x^2 + 6x + 9 = 36(x + 3)$
$x^2 + 6x + 9 = 36x + 108$

Step5: Rearrange to quadratic equation

$x^2 - 30x - 99 = 0$

Step6: Solve quadratic equation

Use quadratic formula $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-30$, $c=-99$:

$$\begin{align*} x&=\frac{30\pm\sqrt{(-30)^2-4(1)(-99)}}{2(1)}\\ &=\frac{30\pm\sqrt{900+396}}{2}\\ &=\frac{30\pm\sqrt{1296}}{2}\\ &=\frac{30\pm36}{2} \end{align*}$$

$x = \frac{30+36}{2}=33$ or $x = \frac{30-36}{2}=-3$

Step7: Verify solutions

• For $x=33$:

$\sqrt{2(33)+15} - \sqrt{33+3} = \sqrt{81} - \sqrt{36} = 9 - 6 = 3$, which matches.

• For $x=-3$:

$\sqrt{2(-3)+15} - \sqrt{-3+3} = \sqrt{9} - \sqrt{0} = 3 - 0 = 3$, which matches.

Answer:

Question 23: $x=-1$
Question 24: $x=33$, $x=-3$