Solutions to rational exponents and radical equations problems, showing step-by-step solutions and identification of extraneous solutions.

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Section III: Solve the following

These are radical and rational exponent equations. We'll solve each one carefully.

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#### a) $\sqrt{4x - 27} - 1 = 4$

Step 1: Isolate the radical
$$
\sqrt{4x - 27} = 5
$$

Step 2: Square both sides
$$
(\sqrt{4x - 27})^2 = 5^2 \Rightarrow 4x - 27 = 25
$$

Step 3: Solve for $x$
$$
4x = 52 \Rightarrow x = 13
$$

✔ Check:
$\sqrt{4(13) - 27} - 1 = \sqrt{52 - 27} - 1 = \sqrt{25} - 1 = 5 - 1 = 4$ ✔️

✔️ Answer: $x = 13$

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#### b) $5\sqrt{x} + 7 = 8$

Step 1: Isolate the radical
$$
5\sqrt{x} = 1 \Rightarrow \sqrt{x} = \frac{1}{5}
$$

Step 2: Square both sides
$$
x = \left(\frac{1}{5}\right)^2 = \frac{1}{25}
$$

✔ Check:
$5\sqrt{\frac{1}{25}} + 7 = 5 \cdot \frac{1}{5} + 7 = 1 + 7 = 8$ ✔️

✔️ Answer: $x = \frac{1}{25}$

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#### c) $2 + (4 - x)^{\frac{3}{2}} = 10$

Step 1: Isolate the power
$$
(4 - x)^{\frac{3}{2}} = 8
$$

Step 2: Raise both sides to the reciprocal power ($\frac{2}{3}$)
$$
\left[(4 - x)^{\frac{3}{2}}\right]^{\frac{2}{3}} = 8^{\frac{2}{3}}
\Rightarrow (4 - x)^1 = (2^3)^{\frac{2}{3}} = 2^2 = 4
$$

So:
$$
4 - x = 4 \Rightarrow x = 0
$$

✔ Check:
$(4 - 0)^{\frac{3}{2}} = 4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$, so $2 + 8 = 10$ ✔️

✔️ Answer: $x = 0$

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#### d) $\sqrt{3x} = \sqrt{x + 4}$

Step 1: Square both sides
$$
(\sqrt{3x})^2 = (\sqrt{x + 4})^2 \Rightarrow 3x = x + 4
$$

Step 2: Solve
$$
2x = 4 \Rightarrow x = 2
$$

✔ Check:
$\sqrt{3(2)} = \sqrt{6}$, $\sqrt{2 + 4} = \sqrt{6}$ → equal ✔️

✔️ Answer: $x = 2$

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#### e) $(x + 4)^{\frac{4}{3}} = 27^{\frac{4}{3}}$

Wait — actually, the equation is:

$$
(x + 4)^{\frac{4}{3}} = 27^{\frac{4}{3}}
$$

But in the image, it says:
$$
(x + 4)^{\frac{4}{3}} = 27^{\frac{4}{3}} \Rightarrow x + 4 = 27^{\frac{4}{3}} \text{?}
$$

Actually, this is not correct unless we're careful.

Let’s solve properly.

We can write:
$$
(x + 4)^{\frac{4}{3}} = 27^{\frac{4}{3}}
\Rightarrow \left[(x + 4)^{\frac{4}{3}}\right] = \left[27^{\frac{4}{3}}\right]
$$

Since both sides have the same exponent, we can equate the bases only if the function is one-to-one. But since the exponent is even in numerator (4), we must consider both positive and negative roots.

But here, let’s take both sides to the power of $\frac{3}{4}$:

$$
x + 4 = \pm 27^{\frac{4}{3} \cdot \frac{3}{4}} = \pm 27^1 = \pm 27
$$

Wait — no! That’s incorrect.

Actually:
$$
\left[(x+4)^{4/3}\right]^{3/4} = (x+4)^1 = \pm 27^{(4/3)(3/4)} = \pm 27^1 = \pm 27
$$

But wait — when you raise a number to the $4/3$ power, it's defined for non-negative numbers in real numbers (since cube root allows negatives, but then raised to 4th power).

But $27^{4/3} = (27^{1/3})^4 = 3^4 = 81$

So:
$$
(x + 4)^{4/3} = 81
$$

Now take both sides to the $3/4$ power:
$$
x + 4 = \pm 81^{3/4}
$$

But $81 = 3^4$, so $81^{3/4} = (3^4)^{3/4} = 3^3 = 27$

So:
$$
x + 4 = \pm 27
\Rightarrow x = 23 \quad \text{or} \quad x = -31
$$

But now check which ones work.

Wait — in the original solution, it says:
$$
x + 4 = 81 \Rightarrow x = 77
$$

That seems wrong.

Wait — let's re-express the problem.

Looking at your image:

> e) $(x + 4)^{4/3} = 27^{4/3}$

But $27^{4/3} = (27^{1/3})^4 = 3^4 = 81$

So:
$$
(x + 4)^{4/3} = 81
$$

Now take both sides to the $3/4$ power:
$$
x + 4 = \pm 81^{3/4}
$$

But $81 = 3^4$, so $81^{3/4} = (3^4)^{3/4} = 3^3 = 27$

So:
$$
x + 4 = \pm 27 \Rightarrow x = 23 \text{ or } x = -31
$$

But in the given solution, they say:
$$
x + 4 = 81 \Rightarrow x = 77
$$

This is incorrect.

Wait — perhaps the original equation was:
$$
(x + 4)^{4/3} = 81
$$

Then:
$$
x + 4 = 81^{3/4} = 27 \Rightarrow x = 23
$$

But even then, $x + 4 = -27$ would give:
$$
(-27)^{4/3} = [(-27)^{1/3}]^4 = (-3)^4 = 81
$$

So both $x = 23$ and $x = -31$ are valid?

Let’s test:

- For $x = 23$: $x + 4 = 27$, $27^{4/3} = (3^3)^{4/3} = 3^4 = 81$ ✔️
- For $x = -31$: $x + 4 = -27$, $(-27)^{4/3} = [(-27)^{1/3}]^4 = (-3)^4 = 81$ ✔️

So both are valid?

But wait — $(-27)^{1/3} = -3$ is fine (cube root of negative is negative), and then $(-3)^4 = 81$, so yes.

So two solutions: $x = 23$ and $x = -31$

But in the image, it says:
$$
x + 4 = 81 \Rightarrow x = 77
$$

That is wrong.

Let me double-check the original problem.

In your image, it says:

> e) $(x + 4)^{4/3} = 27^{4/3}$

But $27^{4/3} = 81$, so:

$$
(x + 4)^{4/3} = 81
\Rightarrow x + 4 = \pm 81^{3/4} = \pm 27
\Rightarrow x = 23 \text{ or } -31
$$

So the solution provided in the image (x = 77) is incorrect.

✘ Error in the given solution.

Correct answer: $x = 23$ or $x = -31$

But let’s see what the image says:

> $x + 4 = 81$ → $x = 77$

That suggests the equation was misread.

Wait — maybe the equation was:

> $(x + 4)^{4/3} = 27^{4/3}$ → but $27^{4/3} = 81$, so $(x+4)^{4/3} = 81$

Then taking both sides to the $3/4$ power:

$$
x + 4 = \pm 81^{3/4} = \pm (3^4)^{3/4} = \pm 3^3 = \pm 27
$$

So $x = 23$ or $x = -31$

But in the image, they wrote:

> $x + 4 = 81$ → $x = 77$

This is not correct.

So either the problem was different, or there’s a typo.

Wait — looking again: the image shows:

> e) $(x + 4)^{4/3} = 27^{4/3}$

But $27^{4/3} = 81$, so the right side is 81.

But they write:

> $x + 4 = 81$ → $x = 77$

Which implies they thought $(x + 4)^{4/3} = 81$ means $x + 4 = 81$, which is false.

They should have done:

$$
(x + 4)^{4/3} = 81
\Rightarrow \text{Raise both sides to } 3/4: (x + 4) = 81^{3/4} = 27
\Rightarrow x = 23
$$

And also consider $x + 4 = -27$ → $x = -31$

But $(-27)^{4/3} = [(-27)^{1/3}]^4 = (-3)^4 = 81$, so it works.

So two solutions: $x = 23$, $x = -31$

But the image only gives $x = 77$, which is wrong.

So the given solution is incorrect.

✔ Correct Answer: $x = 23$ or $x = -31$

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#### f) $\sqrt{(x + 1)^3} - 1 = 7$

Step 1: Isolate radical
$$
\sqrt{(x + 1)^3} = 8
$$

Step 2: Square both sides
$$
(x + 1)^3 = 64
$$

Step 3: Take cube root
$$
x + 1 = \sqrt[3]{64} = 4 \Rightarrow x = 3
$$

✔ Check:
$\sqrt{(3 + 1)^3} - 1 = \sqrt{64} - 1 = 8 - 1 = 7$ ✔️

✔️ Answer: $x = 3$

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Section IV: Solve (Identify any extraneous solutions)

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#### a) $\sqrt{x + 7} + 5 = x$

Step 1: Isolate radical
$$
\sqrt{x + 7} = x - 5
$$

Step 2: Square both sides
$$
x + 7 = (x - 5)^2 = x^2 - 10x + 25
$$

$$
x + 7 = x^2 - 10x + 25
\Rightarrow 0 = x^2 - 11x + 18
\Rightarrow (x - 9)(x - 2) = 0
\Rightarrow x = 9 \text{ or } x = 2
$$

Step 3: Check for extraneous solutions

- $x = 9$: $\sqrt{9 + 7} + 5 = \sqrt{16} + 5 = 4 + 5 = 9$ ✔️
- $x = 2$: $\sqrt{2 + 7} + 5 = \sqrt{9} + 5 = 3 + 5 = 8 \ne 2$ ✘

So $x = 2$ is extraneous.

✔️ Answer: $x = 9$

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#### b) $\sqrt{x + 2} = x$

Step 1: Square both sides
$$
x + 2 = x^2 \Rightarrow x^2 - x - 2 = 0
\Rightarrow (x - 2)(x + 1) = 0 \Rightarrow x = 2 \text{ or } x = -1
$$

Step 2: Check

- $x = 2$: $\sqrt{2 + 2} = \sqrt{4} = 2$ ✔️
- $x = -1$: $\sqrt{-1 + 2} = \sqrt{1} = 1 \ne -1$ ✘

So $x = -1$ is extraneous.

✔️ Answer: $x = 2$

---

#### c) $(5x + 4)^{1/2} - 3x = 0$

$$
\sqrt{5x + 4} = 3x
$$

Square both sides:
$$
5x + 4 = 9x^2
\Rightarrow 9x^2 - 5x - 4 = 0
\Rightarrow (9x + 4)(x - 1) = 0
\Rightarrow x = 1 \text{ or } x = -\frac{4}{9}
$$

Check:

- $x = 1$: $\sqrt{5(1) + 4} = \sqrt{9} = 3$, $3x = 3$ → $3 = 3$ ✔️
- $x = -\frac{4}{9}$: $\sqrt{5(-4/9) + 4} = \sqrt{-20/9 + 36/9} = \sqrt{16/9} = 4/3$, $3x = -12/9 = -4/3$ → $4/3 \ne -4/3$ ✘

So $x = -4/9$ is extraneous.

✔️ Answer: $x = 1$

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#### d) $\sqrt{4x - 5} = 3\sqrt{x - 5}$

Step 1: Square both sides
$$
4x - 5 = 9(x - 5) = 9x - 45
\Rightarrow 4x - 5 = 9x - 45
\Rightarrow -5x = -40 \Rightarrow x = 8
$$

Step 2: Check

- LHS: $\sqrt{4(8) - 5} = \sqrt{32 - 5} = \sqrt{27} = 3\sqrt{3}$
- RHS: $3\sqrt{8 - 5} = 3\sqrt{3}$ → equal ✔️

✔️ Answer: $x = 8$

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#### e) $(x - 9)^{1/2} + 1 = x^{1/2}$

Let’s write as:
$$
\sqrt{x - 9} + 1 = \sqrt{x}
$$

Isolate:
$$
\sqrt{x - 9} = \sqrt{x} - 1
$$

Square both sides:
$$
x - 9 = (\sqrt{x} - 1)^2 = x - 2\sqrt{x} + 1
\Rightarrow x - 9 = x - 2\sqrt{x} + 1
\Rightarrow -9 = -2\sqrt{x} + 1
\Rightarrow -10 = -2\sqrt{x}
\Rightarrow \sqrt{x} = 5 \Rightarrow x = 25
$$

Check:
- LHS: $\sqrt{25 - 9} + 1 = \sqrt{16} + 1 = 4 + 1 = 5$
- RHS: $\sqrt{25} = 5$ → equal ✔️

✔️ Answer: $x = 25$

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#### f) $(x + 5)^{1/2} - (5 - 2x)^{1/4} = 0$

$$
\sqrt{x + 5} = (5 - 2x)^{1/4}
$$

Raise both sides to the 4th power:
$$
(\sqrt{x + 5})^4 = (5 - 2x)
\Rightarrow (x + 5)^2 = 5 - 2x
\Rightarrow x^2 + 10x + 25 = 5 - 2x
\Rightarrow x^2 + 12x + 20 = 0
\Rightarrow (x + 10)(x + 2) = 0
\Rightarrow x = -10 \text{ or } x = -2
$$

Check:

- $x = -10$:
$\sqrt{-10 + 5} = \sqrt{-5}$ → not real ✘
So invalid.

- $x = -2$:
$\sqrt{-2 + 5} = \sqrt{3}$
$(5 - 2(-2))^{1/4} = (5 + 4)^{1/4} = 9^{1/4} = \sqrt{3}$
So $\sqrt{3} = \sqrt{3}$ ✔️

So $x = -2$ is valid.

But $x = -10$ is extraneous.

✔️ Answer: $x = -2$

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✔ Final Summary of Corrections

| Problem | Given Solution | Correct Solution | Notes |
|--------|----------------|------------------|-------|
| III.e) | $x = 77$ | $x = 23$ or $x = -31$ | The solution is incorrect; likely misstep in solving $(x+4)^{4/3} = 81$. Should be $x+4 = \pm 27$ |
| All others | Mostly correct | Correct | Only III.e) has error |

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📌 Key Concepts Used:

- Isolate radicals before squaring.
- Square both sides carefully, and check for extraneous solutions.
- When dealing with even roots, domain restrictions apply (e.g., expressions under even roots must be ≥ 0).
- Rational exponents: $a^{m/n} = (\sqrt[n]{a})^m$, and operations follow exponent rules.

Let me know if you'd like a printable version or explanation on how to avoid common mistakes!