Worksheet with radical equations for algebra practice.

Problem: Solve each radical equation and check for extraneous solutions.

The equations are as follows:

1. \( 10 = \sqrt{5x + 4} \)
2. \( \frac{\sqrt{x}}{3} = 5 \)
3. \( \sqrt{b - 7} + 4 = 8 \)
4. \( 9 + z = \sqrt{9 - 7z} \)
5. \( 6 + \sqrt{10 + x} = 4 \)
6. \( 90 = 10\sqrt{3x} \)
7. \( \sqrt{5a} = \sqrt{6 - a} \)
8. \( \sqrt{u + 3} = \sqrt{3u - 3} \)
9. \( y = \sqrt{-60 + 16y} \)
10. \( w + 6 = \sqrt{72 + w} \)
11. \( 2\sqrt{6m - 35} + 8 = 2m \)
12. \( m + \sqrt{-2m} = 10 \)
13. \( \sqrt{kp - k} - p = 0 \)
14. \( b - \sqrt{b^2 - 4b} = -3 \)

---

Solution:

#### 1. \( 10 = \sqrt{5x + 4} \)

1. Square both sides to eliminate the square root:
\[
10^2 = (\sqrt{5x + 4})^2
\]
\[
100 = 5x + 4
\]

2. Solve for \( x \):
\[
100 - 4 = 5x
\]
\[
96 = 5x
\]
\[
x = \frac{96}{5}
\]

3. Check for extraneous solutions by substituting \( x = \frac{96}{5} \) back into the original equation:
\[
10 = \sqrt{5 \left( \frac{96}{5} \right) + 4}
\]
\[
10 = \sqrt{96 + 4}
\]
\[
10 = \sqrt{100}
\]
\[
10 = 10 \quad \text{(True)}
\]

Therefore, \( x = \frac{96}{5} \) is a valid solution.

\[
\boxed{x = \frac{96}{5}}
\]

---

#### 2. \( \frac{\sqrt{x}}{3} = 5 \)

1. Multiply both sides by 3 to isolate the square root:
\[
\sqrt{x} = 15
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{x})^2 = 15^2
\]
\[
x = 225
\]

3. Check for extraneous solutions by substituting \( x = 225 \) back into the original equation:
\[
\frac{\sqrt{225}}{3} = 5
\]
\[
\frac{15}{3} = 5
\]
\[
5 = 5 \quad \text{(True)}
\]

Therefore, \( x = 225 \) is a valid solution.

\[
\boxed{x = 225}
\]

---

#### 3. \( \sqrt{b - 7} + 4 = 8 \)

1. Isolate the square root term:
\[
\sqrt{b - 7} = 8 - 4
\]
\[
\sqrt{b - 7} = 4
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{b - 7})^2 = 4^2
\]
\[
b - 7 = 16
\]

3. Solve for \( b \):
\[
b = 16 + 7
\]
\[
b = 23
\]

4. Check for extraneous solutions by substituting \( b = 23 \) back into the original equation:
\[
\sqrt{23 - 7} + 4 = 8
\]
\[
\sqrt{16} + 4 = 8
\]
\[
4 + 4 = 8
\]
\[
8 = 8 \quad \text{(True)}
\]

Therefore, \( b = 23 \) is a valid solution.

\[
\boxed{b = 23}
\]

---

#### 4. \( 9 + z = \sqrt{9 - 7z} \)

1. Isolate the square root term:
\[
\sqrt{9 - 7z} = 9 + z
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{9 - 7z})^2 = (9 + z)^2
\]
\[
9 - 7z = 81 + 18z + z^2
\]

3. Rearrange into standard quadratic form:
\[
z^2 + 18z + 7z + 81 - 9 = 0
\]
\[
z^2 + 25z + 72 = 0
\]

4. Factor the quadratic equation:
\[
(z + 8)(z + 9) = 0
\]

5. Solve for \( z \):
\[
z + 8 = 0 \quad \text{or} \quad z + 9 = 0
\]
\[
z = -8 \quad \text{or} \quad z = -9
\]

6. Check for extraneous solutions by substituting \( z = -8 \) and \( z = -9 \) back into the original equation:

- For \( z = -8 \):
\[
9 + (-8) = \sqrt{9 - 7(-8)}
\]
\[
1 = \sqrt{9 + 56}
\]
\[
1 = \sqrt{65} \quad \text{(False)}
\]

- For \( z = -9 \):
\[
9 + (-9) = \sqrt{9 - 7(-9)}
\]
\[
0 = \sqrt{9 + 63}
\]
\[
0 = \sqrt{72} \quad \text{(False)}
\]

Both solutions are extraneous.

\[
\boxed{\text{No solution}}
\]

---

#### 5. \( 6 + \sqrt{10 + x} = 4 \)

1. Isolate the square root term:
\[
\sqrt{10 + x} = 4 - 6
\]
\[
\sqrt{10 + x} = -2
\]

2. Note that the square root of a real number cannot be negative. Therefore, there is no real solution.

\[
\boxed{\text{No solution}}
\]

---

#### 6. \( 90 = 10\sqrt{3x} \)

1. Divide both sides by 10 to isolate the square root:
\[
\sqrt{3x} = 9
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{3x})^2 = 9^2
\]
\[
3x = 81
\]

3. Solve for \( x \):
\[
x = \frac{81}{3}
\]
\[
x = 27
\]

4. Check for extraneous solutions by substituting \( x = 27 \) back into the original equation:
\[
90 = 10\sqrt{3(27)}
\]
\[
90 = 10\sqrt{81}
\]
\[
90 = 10 \cdot 9
\]
\[
90 = 90 \quad \text{(True)}
\]

Therefore, \( x = 27 \) is a valid solution.

\[
\boxed{x = 27}
\]

---

#### 7. \( \sqrt{5a} = \sqrt{6 - a} \)

1. Square both sides to eliminate the square roots:
\[
(\sqrt{5a})^2 = (\sqrt{6 - a})^2
\]
\[
5a = 6 - a
\]

2. Solve for \( a \):
\[
5a + a = 6
\]
\[
6a = 6
\]
\[
a = 1
\]

3. Check for extraneous solutions by substituting \( a = 1 \) back into the original equation:
\[
\sqrt{5(1)} = \sqrt{6 - 1}
\]
\[
\sqrt{5} = \sqrt{5} \quad \text{(True)}
\]

Therefore, \( a = 1 \) is a valid solution.

\[
\boxed{a = 1}
\]

---

#### 8. \( \sqrt{u + 3} = \sqrt{3u - 3} \)

1. Square both sides to eliminate the square roots:
\[
(\sqrt{u + 3})^2 = (\sqrt{3u - 3})^2
\]
\[
u + 3 = 3u - 3
\]

2. Solve for \( u \):
\[
u - 3u = -3 - 3
\]
\[
-2u = -6
\]
\[
u = 3
\]

3. Check for extraneous solutions by substituting \( u = 3 \) back into the original equation:
\[
\sqrt{3 + 3} = \sqrt{3(3) - 3}
\]
\[
\sqrt{6} = \sqrt{9 - 3}
\]
\[
\sqrt{6} = \sqrt{6} \quad \text{(True)}
\]

Therefore, \( u = 3 \) is a valid solution.

\[
\boxed{u = 3}
\]

---

#### 9. \( y = \sqrt{-60 + 16y} \)

1. Square both sides to eliminate the square root:
\[
y^2 = (\sqrt{-60 + 16y})^2
\]
\[
y^2 = -60 + 16y
\]

2. Rearrange into standard quadratic form:
\[
y^2 - 16y + 60 = 0
\]

3. Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 1, \, b = -16, \, c = 60
\]
\[
y = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(60)}}{2(1)}
\]
\[
y = \frac{16 \pm \sqrt{256 - 240}}{2}
\]
\[
y = \frac{16 \pm \sqrt{16}}{2}
\]
\[
y = \frac{16 \pm 4}{2}
\]

4. Solve for the two possible values of \( y \):
\[
y = \frac{16 + 4}{2} = \frac{20}{2} = 10
\]
\[
y = \frac{16 - 4}{2} = \frac{12}{2} = 6
\]

5. Check for extraneous solutions by substituting \( y = 10 \) and \( y = 6 \) back into the original equation:

- For \( y = 10 \):
\[
10 = \sqrt{-60 + 16(10)}
\]
\[
10 = \sqrt{-60 + 160}
\]
\[
10 = \sqrt{100}
\]
\[
10 = 10 \quad \text{(True)}
\]

- For \( y = 6 \):
\[
6 = \sqrt{-60 + 16(6)}
\]
\[
6 = \sqrt{-60 + 96}
\]
\[
6 = \sqrt{36}
\]
\[
6 = 6 \quad \text{(True)}
\]

Both solutions are valid.

\[
\boxed{y = 10, \, y = 6}
\]

---

#### 10. \( w + 6 = \sqrt{72 + w} \)

1. Isolate the square root term:
\[
\sqrt{72 + w} = w + 6
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{72 + w})^2 = (w + 6)^2
\]
\[
72 + w = w^2 + 12w + 36
\]

3. Rearrange into standard quadratic form:
\[
w^2 + 12w + 36 - w - 72 = 0
\]
\[
w^2 + 11w - 36 = 0
\]

4. Factor the quadratic equation:
\[
(w + 12)(w - 3) = 0
\]

5. Solve for \( w \):
\[
w + 12 = 0 \quad \text{or} \quad w - 3 = 0
\]
\[
w = -12 \quad \text{or} \quad w = 3
\]

6. Check for extraneous solutions by substituting \( w = -12 \) and \( w = 3 \) back into the original equation:

- For \( w = -12 \):
\[
-12 + 6 = \sqrt{72 + (-12)}
\]
\[
-6 = \sqrt{60} \quad \text{(False)}
\]

- For \( w = 3 \):
\[
3 + 6 = \sqrt{72 + 3}
\]
\[
9 = \sqrt{75} \quad \text{(False)}
\]

Both solutions are extraneous.

\[
\boxed{\text{No solution}}
\]

---

#### 11. \( 2\sqrt{6m - 35} + 8 = 2m \)

1. Isolate the square root term:
\[
2\sqrt{6m - 35} = 2m - 8
\]
\[
\sqrt{6m - 35} = m - 4
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{6m - 35})^2 = (m - 4)^2
\]
\[
6m - 35 = m^2 - 8m + 16
\]

3. Rearrange into standard quadratic form:
\[
m^2 - 8m + 16 - 6m + 35 = 0
\]
\[
m^2 - 14m + 51 = 0
\]

4. Solve the quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 1, \, b = -14, \, c = 51
\]
\[
m = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(51)}}{2(1)}
\]
\[
m = \frac{14 \pm \sqrt{196 - 204}}{2}
\]
\[
m = \frac{14 \pm \sqrt{-8}}{2}
\]

5. Since the discriminant is negative (\( \sqrt{-8} \)), there are no real solutions.

\[
\boxed{\text{No solution}}
\]

---

#### 12. \( m + \sqrt{-2m} = 10 \)

1. Isolate the square root term:
\[
\sqrt{-2m} = 10 - m
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{-2m})^2 = (10 - m)^2
\]
\[
-2m = 100 - 20m + m^2
\]

3. Rearrange into standard quadratic form:
\[
m^2 - 20m + 100 + 2m = 0
\]
\[
m^2 - 18m + 100 = 0
\]

4. Solve the quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 1, \, b = -18, \, c = 100
\]
\[
m = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(100)}}{2(1)}
\]
\[
m = \frac{18 \pm \sqrt{324 - 400}}{2}
\]
\[
m = \frac{18 \pm \sqrt{-76}}{2}
\]

5. Since the discriminant is negative (\( \sqrt{-76} \)), there are no real solutions.

\[
\boxed{\text{No solution}}
\]

---

#### 13. \( \sqrt{kp - k} - p = 0 \)

1. Isolate the square root term:
\[
\sqrt{kp - k} = p
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{kp - k})^2 = p^2
\]
\[
kp - k = p^2
\]

3. Factor out \( k \) on the left-hand side:
\[
k(p - 1) = p^2
\]

4. Solve for \( k \):
\[
k = \frac{p^2}{p - 1}
\]

5. Check for extraneous solutions. The expression \( k = \frac{p^2}{p - 1} \) is valid as long as \( p \neq 1 \).

\[
\boxed{k = \frac{p^2}{p - 1}}
\]

---

#### 14. \( b - \sqrt{b^2 - 4b} = -3 \)

1. Isolate the square root term:
\[
\sqrt{b^2 - 4b} = b + 3
\]

2. Square both sides to eliminate the square root:
\[
(\sqrt{b^2 - 4b})^2 = (b + 3)^2
\]
\[
b^2 - 4b = b^2 + 6b + 9
\]

3. Simplify and solve for \( b \):
\[
b^2 - 4b - b^2 - 6b - 9 = 0
\]
\[
-10b - 9 = 0
\]
\[
-10b = 9
\]
\[
b = -\frac{9}{10}
\]

4. Check for extraneous solutions by substituting \( b = -\frac{9}{10} \) back into the original equation:
\[
-\frac{9}{10} - \sqrt{\left( -\frac{9}{10} \right)^2 - 4\left( -\frac{9}{10} \right)} = -3
\]
\[
-\frac{9}{10} - \sqrt{\frac{81}{100} + \frac{36}{10}} = -3
\]
\[
-\frac{9}{10} - \sqrt{\frac{81}{100} + \frac{360}{100}} = -3
\]
\[
-\frac{9}{10} - \sqrt{\frac{441}{100}} = -3
\]
\[
-\frac{9}{10} - \frac{21}{10} = -3
\]
\[
-\frac{30}{10} = -3
\]
\[
-3 = -3 \quad \text{(True)}
\]

Therefore, \( b = -\frac{9}{10} \) is a valid solution.

\[
\boxed{b = -\frac{9}{10}}
\]

---

Final Answers:

1. \( \boxed{x = \frac{96}{5}} \)
2. \( \boxed{x = 225} \)
3. \( \boxed{b = 23} \)
4. \( \boxed{\text{No solution}} \)
5. \( \boxed{\text{No solution}} \)
6. \( \boxed{x = 27} \)
7. \( \boxed{a = 1} \)
8. \( \boxed{u = 3} \)
9. \( \boxed{y = 10, \, y = 6} \)
10. \( \boxed{\text{No solution}} \)
11. \( \boxed{\text{No solution}} \)
12. \( \boxed{\text{No solution}} \)
13. \( \boxed{k = \frac{p^2}{p - 1}} \)
14. \( \boxed{b = -\frac{9}{10}} \)