in exercises 27 - 34, solve the equation. check your solution(s). (see examples 5 and 6.)
27. $2x^{3/3}=8$
28. $4x^{3/2}=32$
29. $x^{1/4}+3 = 0$
30. $2x^{3/6}-14 = 40$
31. $(x + 6)^{1/2}=x$
32. $(5 - x)^{1/2}-2x = 0$
33. $2(x + 11)^{1/2}=x + 3$
34. $(5x^{2}-4)^{1/4}=x$

Question

Accepted Answer

### Problem 27: Solve $2x^{\frac{2}{3}} = 8$
# Explanation:
## Step 1: Isolate the variable term
Divide both sides by 2:
$x^{\frac{2}{3}}=\frac{8}{2} = 4$

## Step 2: Eliminate the fractional exponent
Raise both sides to the power of $\frac{3}{2}$ (since $\frac{2}{3}\times\frac{3}{2}=1$):
$x = 4^{\frac{3}{2}}$

## Step 3: Simplify $4^{\frac{3}{2}}$
Recall $a^{\frac{m}{n}}=\sqrt[n]{a^m}$ or $(\sqrt[n]{a})^m$. Here, $4^{\frac{3}{2}}=(\sqrt{4})^3 = 2^3=8$

## Step 4: Check the solution
Substitute $x = 8$ into the original equation:
Left - hand side: $2\times(8)^{\frac{2}{3}}=2\times((8)^{\frac{1}{3}})^2=2\times(2)^2 = 2\times4 = 8$
Right - hand side: $8$
Since LHS = RHS, $x = 8$ is a valid solution.

# Answer: $x = 8$

### Problem 28: Solve $4x^{\frac{3}{2}} = 32$
# Explanation:
## Step 1: Isolate the variable term
Divide both sides by 4:
$x^{\frac{3}{2}}=\frac{32}{4}=8$

## Step 2: Eliminate the fractional exponent
Raise both sides to the power of $\frac{2}{3}$:
$x = 8^{\frac{2}{3}}$

## Step 3: Simplify $8^{\frac{2}{3}}$
$8^{\frac{2}{3}}=(\sqrt[3]{8})^2=2^2 = 4$

## Step 4: Check the solution
Substitute $x = 4$ into the original equation:
Left - hand side: $4\times(4)^{\frac{3}{2}}=4\times((4)^{\frac{1}{2}})^3=4\times(2)^3=4\times8 = 32$
Right - hand side: $32$
Since LHS = RHS, $x = 4$ is a valid solution.

# Answer: $x = 4$

### Problem 29: Solve $x^{\frac{1}{4}}+3 = 0$
# Explanation:
## Step 1: Isolate the variable term
Subtract 3 from both sides:
$x^{\frac{1}{4}}=- 3$

## Step 2: Analyze the domain
The expression $x^{\frac{1}{4}}=\sqrt[4]{x}$ is defined only for $x\geq0$ (since the fourth - root of a negative number is not a real number), and $\sqrt[4]{x}\geq0$ for all $x\geq0$. But we have $\sqrt[4]{x}=-3$, which is impossible because the left - hand side is non - negative and the right - hand side is negative.

# Answer: No real solution

### Problem 30: Solve $2x^{\frac{3}{4}}-14 = 40$
# Explanation:
## Step 1: Isolate the variable term
Add 14 to both sides:
$2x^{\frac{3}{4}}=40 + 14=54$

## Step 2: Isolate $x^{\frac{3}{4}}$
Divide both sides by 2:
$x^{\frac{3}{4}}=\frac{54}{2}=27$

## Step 3: Eliminate the fractional exponent
Raise both sides to the power of $\frac{4}{3}$:
$x = 27^{\frac{4}{3}}$

## Step 4: Simplify $27^{\frac{4}{3}}$
$27^{\frac{4}{3}}=(\sqrt[3]{27})^4=3^4 = 81$

## Step 5: Check the solution
Substitute $x = 81$ into the original equation:
Left - hand side: $2\times(81)^{\frac{3}{4}}-14=2\times((81)^{\frac{1}{4}})^3-14=2\times(3)^3-14=2\times27 - 14=54 - 14 = 40$
Right - hand side: $40$
Since LHS = RHS, $x = 81$ is a valid solution.

# Answer: $x = 81$

### Problem 31: Solve $(x + 6)^{\frac{1}{2}}=x$
# Explanation:
## Step 1: Square both sides to eliminate the square root
$(x + 6)=x^2$ (note that $x\geq0$ because the left - hand side $\sqrt{x + 6}\geq0$, so $x\geq0$)

## Step 2: Rearrange into a quadratic equation
$x^2-x - 6=0$

## Step 3: Factor the quadratic equation
$x^2-x - 6=(x - 3)(x+2)=0$

## Step 4: Solve for $x$
$x - 3=0$ or $x + 2=0$, so $x = 3$ or $x=-2$

## Step 5: Check the solutions
- For $x = 3$: Left - hand side $\sqrt{3 + 6}=\sqrt{9}=3$, Right - hand side $=3$. So $x = 3$ is valid.
- For $x=-2$: Left - hand side $\sqrt{-2 + 6}=\sqrt{4}=2$, Right - hand side $=-2$. Since $2
eq - 2$, $x=-2$ is not valid (because we squared both sides, we may have introduced an extraneous solution).

# Answer: $x = 3$

### Problem 32: Solve $(5 - x)^{\frac{1}{2}}-2x = 0$
# Explanation:
## Step 1: Isolate the square root term
$(5 - x)^{\frac{1}{2}}=2x$ (note that $5 - x\geq0$ (so $x\leq5$) and $2x\geq0$ (so $x\geq0$), so $0\leq x\leq5$)

## Step 2: Square both sides
$5 - x = 4x^2$

## Step 3: Rearrange into a quadratic equation
$4x^2+x - 5=0$

## Step 4: Factor the quadratic equation
$4x^2+x - 5=(4x + 5)(x - 1)=0$

## Step 5: Solve for $x$
$4x+5 = 0$ or $x - 1=0$, so $x=-\frac{5}{4}$ or $x = 1$

## Step 6: Check the solutions
- For $x = 1$: Left - hand side $\sqrt{5 - 1}-2\times1=\sqrt{4}-2=2 - 2 = 0$, Right - hand side $=0$. So $x = 1$ is valid.
- For $x=-\frac{5}{4}$: Left - hand side $\sqrt{5+\frac{5}{4}}-2\times(-\frac{5}{4})=\sqrt{\frac{25}{4}}+\frac{5}{2}=\frac{5}{2}+\frac{5}{2}=5
eq0$. Also, $x =-\frac{5}{4}<0$ and $2x=-\frac{5}{2}<0$, but $\sqrt{5 - x}\geq0$, so $x=-\frac{5}{4}$ is not valid.

# Answer: $x = 1$

### Problem 33: Solve $2(x + 11)^{\frac{1}{2}}=x + 3$
# Explanation:
## Step 1: Isolate the square root term
Divide both sides by 2:
$(x + 11)^{\frac{1}{2}}=\frac{x + 3}{2}$ (note that $x+11\geq0$ (so $x\geq - 11$) and $\frac{x + 3}{2}\geq0$ (so $x\geq - 3$), so $x\geq - 3$)

## Step 2: Square both sides
$x + 11=\frac{(x + 3)^2}{4}$

## Step 3: Multiply both sides by 4 to eliminate the denominator
$4(x + 11)=(x + 3)^2$

## Step 4: Expand and rearrange
$4x+44=x^2 + 6x + 9$
$x^2+2x - 35 = 0$

## Step 5: Factor the quadratic equation
$x^2+2x - 35=(x + 7)(x - 5)=0$

## Step 6: Solve for $x$
$x=-7$ or $x = 5$

## Step 7: Check the solutions
- For $x = 5$: Left - hand side $2\sqrt{5 + 11}=2\sqrt{16}=2\times4 = 8$, Right - hand side $=5 + 3 = 8$. So $x = 5$ is valid.
- For $x=-7$: Left - hand side $2\sqrt{-7 + 11}=2\sqrt{4}=4$, Right - hand side $=-7 + 3=-4$. Since $4
eq - 4$, $x=-7$ is not valid (because $x=-7<-3$ which violates the domain condition from step 1).

# Answer: $x = 5$

### Problem 34: Solve $(5x^2-4)^{\frac{1}{4}}=x$
# Explanation:
## Step 1: Raise both sides to the power of 4 to eliminate the fourth root
$5x^2-4=x^4$

## Step 2: Rearrange into a quartic equation
$x^4-5x^2 + 4=0$

## Step 3: Factor the quartic equation (let $y = x^2$)
$y^2-5y + 4=0$, where $y=x^2$
$(y - 1)(y - 4)=0$

## Step 4: Substitute back $y = x^2$ and solve for $x$
- If $y - 1=0$, then $x^2=1$, so $x=\pm1$
- If $y - 4=0$, then $x^2=4$, so $x=\pm2$

## Step 5: Check the solutions
- For $x = 1$: Left - hand side $\sqrt[4]{5\times1^2-4}=\sqrt[4]{1}=1$, Right - hand side $=1$. Valid.
- For $x=-1$: Left - hand side $\sqrt[4]{5\times(-1)^2-4}=\sqrt[4]{1}=1$, Right - hand side $=-1$. Since $1
eq - 1$, invalid.
- For $x = 2$: Left - hand side $\sqrt[4]{5\times2^2-4}=\sqrt[4]{16}=2$, Right - hand side $=2$. Valid.
- For $x=-2$: Left - hand side $\sqrt[4]{5\times(-2)^2-4}=\sqrt[4]{16}=2$, Right - hand side $=-2$. Since $2
eq - 2$, invalid.

# Answer: $x = 1$ or $x = 2$

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QUESTION IMAGE

Question

Explanation:

Problem 27: Solve \(2x^{\frac{2}{3}} = 8\)

Step 1: Isolate the variable term

Step 2: Eliminate the fractional exponent

Step 3: Simplify \(4^{\frac{3}{2}}\)

Step 4: Check the solution

Step 1: Isolate the variable term

Step 2: Eliminate the fractional exponent

Step 3: Simplify \(8^{\frac{2}{3}}\)

Step 4: Check the solution

Step 1: Isolate the variable term

Step 2: Analyze the domain

Answer:

Problem 28: Solve \(4x^{\frac{3}{2}} = 32\)