Question

1. $(6x + 8)^{1/2} - 3x = 0$
2. $(2x^{2} + 8)^{1/4} = x$
3. $2x^{2/3} = 18$
4. $x^{3/4} + 10 = 0$
5. $(x + 12)^{1/2} = x$

Explanation:

Problem 14: $(6x + 8)^{1/2} - 3x = 0$

Step1: Isolate the radical term

$(6x + 8)^{1/2} = 3x$

Step2: Square both sides

$(6x + 8) = (3x)^2$
$6x + 8 = 9x^2$

Step3: Rearrange to quadratic form

$9x^2 - 6x - 8 = 0$

Step4: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=9, b=-6, c=-8$:
$\Delta=(-6)^2-4(9)(-8)=36+288=324$
$x=\frac{6\pm\sqrt{324}}{18}=\frac{6\pm18}{18}$

Step5: Check valid solutions

$x=\frac{6+18}{18}=\frac{24}{18}=\frac{4}{3}$; $x=\frac{6-18}{18}=\frac{-12}{18}=-\frac{2}{3}$
Substitute $x=-\frac{2}{3}$ into original equation: $\sqrt{6(-\frac{2}{3})+8}-3(-\frac{2}{3})=\sqrt{4}+2=4
eq0$ (invalid)

Step1: Raise both sides to 4th power

$2x^2 + 8 = x^4$

Step2: Rearrange to quartic form

$x^4 - 2x^2 - 8 = 0$

Step3: Substitute $u=x^2$

$u^2 - 2u - 8 = 0$

Step4: Solve quadratic for $u$

$(u-4)(u+2)=0$ so $u=4$ or $u=-2$

Step5: Solve for $x$ and check validity

$u=x^2=4\Rightarrow x=\pm2$; $u=x^2=-2$ (no real solution)
Substitute $x=-2$ into original equation: $(2(4)+8)^{1/4}=(16)^{1/4}=2
eq-2$ (invalid)

Step1: Isolate the variable term

$x^{2/3} = 9$

Step2: Raise both sides to 3/2 power

$x = 9^{3/2}=(9^{1/2})^3$

Step3: Calculate the value

$x=3^3=27$; also $x=(-3)^3=-27$ (since even exponent in root)

Answer:

$x=\frac{4}{3}$

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Problem 15: $(2x^2 + 8)^{1/4} = x$