Solve radical equations to find the value of the unknown variable.

Let's solve each of the radical equations step by step. Radical equations involve variables under a radical (square root, cube root, etc.). The key strategy is to isolate the radical, then square (or cube, etc.) both sides to eliminate it, and finally check for extraneous solutions.

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1. $\sqrt{p + 9} = 1$

Step 1: Square both sides:
$$
(\sqrt{p + 9})^2 = 1^2 \Rightarrow p + 9 = 1
$$

Step 2: Solve for $p$:
$$
p = 1 - 9 = -8
$$

Check:
$$
\sqrt{-8 + 9} = \sqrt{1} = 1 \quad \text{✓ Valid}
$$

✔ Answer: $p = -8$

---

2. $10\sqrt{9g} = 60$

Step 1: Divide both sides by 10:
$$
\sqrt{9g} = 6
$$

Step 2: Square both sides:
$$
9g = 36
$$

Step 3: Solve for $g$:
$$
g = \frac{36}{9} = 4
$$

Check:
$$
10\sqrt{9 \cdot 4} = 10\sqrt{36} = 10 \cdot 6 = 60 \quad \text{✓ Valid}
$$

✔ Answer: $g = 4$

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3. $\sqrt{x} + \sqrt{x + 11} = 11$

Let’s isolate one radical:
$$
\sqrt{x + 11} = 11 - \sqrt{x}
$$

Step 1: Square both sides:
$$
(\sqrt{x + 11})^2 = (11 - \sqrt{x})^2
\Rightarrow x + 11 = 121 - 22\sqrt{x} + x
$$

Step 2: Simplify:
$$
x + 11 = 121 + x - 22\sqrt{x}
$$

Subtract $x$ from both sides:
$$
11 = 121 - 22\sqrt{x}
$$

Step 3: Solve:
$$
-110 = -22\sqrt{x} \Rightarrow \sqrt{x} = 5
$$

Step 4: Square both sides:
$$
x = 25
$$

Check:
$$
\sqrt{25} + \sqrt{25 + 11} = 5 + \sqrt{36} = 5 + 6 = 11 \quad \text{✓ Valid}
$$

✔ Answer: $x = 25$

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4. $\sqrt[4]{2p + 2} = 3$

This is a fourth root, so raise both sides to the 4th power:
$$
(\sqrt[4]{2p + 2})^4 = 3^4 \Rightarrow 2p + 2 = 81
$$

Step 2: Solve:
$$
2p = 79 \Rightarrow p = \frac{79}{2} = 39.5
$$

Check:
$$
\sqrt[4]{2(39.5) + 2} = \sqrt[4]{79 + 2} = \sqrt[4]{81} = 3 \quad \text{✓ Valid}
$$

✔ Answer: $p = \frac{79}{2}$ or $39.5$

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

Step 1: Isolate the radical:
$$
\sqrt{2 - x} = x + 7 - 3 = x + 4
$$

Step 2: Square both sides:
$$
(\sqrt{2 - x})^2 = (x + 4)^2 \Rightarrow 2 - x = x^2 + 8x + 16
$$

Step 3: Bring all terms to one side:
$$
0 = x^2 + 8x + 16 + x - 2 = x^2 + 9x + 14
$$

Factor:
$$
x^2 + 9x + 14 = (x + 7)(x + 2) = 0
\Rightarrow x = -7 \text{ or } x = -2
$$

Check both:

- $x = -7$:
$\sqrt{2 - (-7)} + 3 = \sqrt{9} + 3 = 3 + 3 = 6$
Right side: $-7 + 7 = 0$ → $6 \ne 0$ ✘ Invalid

- $x = -2$:
$\sqrt{2 - (-2)} + 3 = \sqrt{4} + 3 = 2 + 3 = 5$
Right side: $-2 + 7 = 5$ → ✓ Valid

✔ Answer: $x = -2$

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6. $\sqrt{2b + 3} = \sqrt{5b - 3}$

Step 1: Square both sides:
$$
(\sqrt{2b + 3})^2 = (\sqrt{5b - 3})^2 \Rightarrow 2b + 3 = 5b - 3
$$

Step 2: Solve:
$$
3 + 3 = 5b - 2b \Rightarrow 6 = 3b \Rightarrow b = 2
$$

Check:
$$
\sqrt{2(2) + 3} = \sqrt{7}, \quad \sqrt{5(2) - 3} = \sqrt{7} \quad \text{✓ Valid}
$$

✔ Answer: $b = 2$

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7. $2f = \sqrt{4f^2 + 6f - 12}$

Step 1: Square both sides:
$$
(2f)^2 = (\sqrt{4f^2 + 6f - 12})^2 \Rightarrow 4f^2 = 4f^2 + 6f - 12
$$

Step 2: Subtract $4f^2$ from both sides:
$$
0 = 6f - 12 \Rightarrow 6f = 12 \Rightarrow f = 2
$$

Check:
Left: $2(2) = 4$
Right: $\sqrt{4(4) + 6(2) - 12} = \sqrt{16 + 12 - 12} = \sqrt{16} = 4$ → ✓ Valid

✔ Answer: $f = 2$

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8. $\sqrt{2t + 15} = t$

Step 1: Square both sides:
$$
(\sqrt{2t + 15})^2 = t^2 \Rightarrow 2t + 15 = t^2
$$

Step 2: Rearrange:
$$
t^2 - 2t - 15 = 0
$$

Factor:
$$
(t - 5)(t + 3) = 0 \Rightarrow t = 5 \text{ or } t = -3
$$

Check:

- $t = 5$: $\sqrt{2(5) + 15} = \sqrt{10 + 15} = \sqrt{25} = 5$ → ✓ Valid
- $t = -3$: $\sqrt{2(-3) + 15} = \sqrt{-6 + 15} = \sqrt{9} = 3$, but right side is $-3$ → $3 \ne -3$ ✘ Invalid

✔ Answer: $t = 5$

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9. $2\sqrt{x + 1} - \sqrt{2x} = \sqrt{x - 4}$

This one is more complex. Let’s try to simplify and check possible values.

Let’s assume $x$ is such that all radicals are defined:

- $x + 1 \geq 0 \Rightarrow x \geq -1$
- $2x \geq 0 \Rightarrow x \geq 0$
- $x - 4 \geq 0 \Rightarrow x \geq 4$

So domain: $x \geq 4$

Try $x = 4$:
- LHS: $2\sqrt{5} - \sqrt{8} \approx 2(2.236) - 2.828 = 4.472 - 2.828 = 1.644$
- RHS: $\sqrt{0} = 0$ → Not equal

Try $x = 5$:
- LHS: $2\sqrt{6} - \sqrt{10} \approx 2(2.45) - 3.16 = 4.9 - 3.16 = 1.74$
- RHS: $\sqrt{1} = 1$ → No

Try $x = 9$:
- LHS: $2\sqrt{10} - \sqrt{18} \approx 2(3.16) - 4.24 = 6.32 - 4.24 = 2.08$
- RHS: $\sqrt{5} \approx 2.236$ → Close but not equal

Try $x = 16$:
- LHS: $2\sqrt{17} - \sqrt{32} \approx 2(4.123) - 5.657 = 8.246 - 5.657 = 2.589$
- RHS: $\sqrt{12} \approx 3.464$ → No

Try $x = 25$:
- LHS: $2\sqrt{26} - \sqrt{50} \approx 2(5.1) - 7.07 = 10.2 - 7.07 = 3.13$
- RHS: $\sqrt{21} \approx 4.58$ → No

Wait — maybe try squaring both sides carefully.

Let:
$$
2\sqrt{x+1} - \sqrt{2x} = \sqrt{x - 4}
$$

Isolate one radical. Move $\sqrt{2x}$ to the right:
$$
2\sqrt{x+1} = \sqrt{x - 4} + \sqrt{2x}
$$

Now square both sides:
$$
(2\sqrt{x+1})^2 = (\sqrt{x - 4} + \sqrt{2x})^2
\Rightarrow 4(x+1) = (x - 4) + 2\sqrt{(x - 4)(2x)} + 2x
$$

Simplify:
$$
4x + 4 = x - 4 + 2x + 2\sqrt{2x(x - 4)}
\Rightarrow 4x + 4 = 3x - 4 + 2\sqrt{2x(x - 4)}
$$

Subtract $3x - 4$:
$$
x + 8 = 2\sqrt{2x(x - 4)}
$$

Divide both sides by 2:
$$
\frac{x + 8}{2} = \sqrt{2x(x - 4)}
$$

Now square again:
$$
\left(\frac{x + 8}{2}\right)^2 = 2x(x - 4)
\Rightarrow \frac{(x + 8)^2}{4} = 2x^2 - 8x
$$

Multiply both sides by 4:
$$
(x + 8)^2 = 8x^2 - 32x
\Rightarrow x^2 + 16x + 64 = 8x^2 - 32x
$$

Bring all terms to one side:
$$
0 = 7x^2 - 48x - 64
$$

Use quadratic formula:
$$
x = \frac{48 \pm \sqrt{(-48)^2 - 4(7)(-64)}}{2(7)} = \frac{48 \pm \sqrt{2304 + 1792}}{14} = \frac{48 \pm \sqrt{4096}}{14}
$$

$\sqrt{4096} = 64$

$$
x = \frac{48 \pm 64}{14}
\Rightarrow x = \frac{112}{14} = 8 \quad \text{or} \quad x = \frac{-16}{14} = -\frac{8}{7}
$$

But $x \geq 4$, so only $x = 8$ is in domain.

Check $x = 8$:

LHS: $2\sqrt{9} - \sqrt{16} = 2(3) - 4 = 6 - 4 = 2$

RHS: $\sqrt{8 - 4} = \sqrt{4} = 2$ → ✓ Valid

✔ Answer: $x = 8$

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10. $10 + \sqrt{10m - 1} = 13$

Step 1: Isolate radical:
$$
\sqrt{10m - 1} = 13 - 10 = 3
$$

Step 2: Square both sides:
$$
10m - 1 = 9 \Rightarrow 10m = 10 \Rightarrow m = 1
$$

Check:
$$
10 + \sqrt{10(1) - 1} = 10 + \sqrt{9} = 10 + 3 = 13 \quad \text{✓ Valid}
$$

✔ Answer: $m = 1$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $p = -8$ |
| 2 | $g = 4$ |
| 3 | $x = 25$ |
| 4 | $p = \frac{79}{2}$ |
| 5 | $x = -2$ |
| 6 | $b = 2$ |
| 7 | $f = 2$ |
| 8 | $t = 5$ |
| 9 | $x = 8$ |
| 10 | $m = 1$ |

Let me know if you'd like these formatted as a printable answer sheet!