Worksheet 3.6: Solve Radical Equations Practice B

Let’s solve each equation step by step. We’ll go one at a time, checking our work as we go.

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1. √x + 3 = 12

Subtract 3 from both sides:
√x = 9

Square both sides:
x = 81

Check: √81 + 3 = 9 + 3 = 12 → ✔️ Correct.

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2. x^(1/2) - 4 = 1

Add 4 to both sides:
x^(1/2) = 5

That means √x = 5 → square both sides:
x = 25

Check: √25 - 4 = 5 - 4 = 1 → ✔️ Correct.

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3. 3√(x + 2) = 6

Divide both sides by 3:
√(x + 2) = 2

Square both sides:
x + 2 = 4 → x = 2

Check: 3√(2+2) = 3√4 = 3×2 = 6 → ✔️ Correct.

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4. (2x - 3)^(1/2) + 2 = 2

Subtract 2 from both sides:
(2x - 3)^(1/2) = 0 → √(2x - 3) = 0

Square both sides:
2x - 3 = 0 → 2x = 3 → x = 3/2

Check: √(2*(3/2) - 3) + 2 = √(3 - 3) + 2 = 0 + 2 = 2 → ✔️ Correct.

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5. 5√(3x) = 15

Divide both sides by 5:
√(3x) = 3

Square both sides:
3x = 9 → x = 3

Check: 5√(3*3) = 5√9 = 5×3 = 15 → ✔️ Correct.

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6. 3√(4 - 3x) = 21

Divide both sides by 3:
√(4 - 3x) = 7

Square both sides:
4 - 3x = 49 → -3x = 45 → x = -15

Check: 3√(4 - 3*(-15)) = 3√(4 + 45) = 3√49 = 3×7 = 21 → ✔️ Correct.

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7. 7 - √(x - 4) = -6

Subtract 7 from both sides:
-√(x - 4) = -13

Multiply both sides by -1:
√(x - 4) = 13

Square both sides:
x - 4 = 169 → x = 173

Check: 7 - √(173 - 4) = 7 - √169 = 7 - 13 = -6 → ✔️ Correct.

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8. √(3x + 4) + 3/2 = 3

Subtract 3/2 from both sides:
√(3x + 4) = 3 - 3/2 = 3/2

Square both sides:
3x + 4 = (3/2)^2 = 9/4

Subtract 4:
3x = 9/4 - 16/4 = -7/4

Divide by 3:
x = (-7/4)/3 = -7/12

Check: √(3*(-7/12) + 4) + 3/2 = √(-21/12 + 48/12) + 3/2 = √(27/12) + 3/2 = √(9/4) + 3/2 = 3/2 + 3/2 = 3 → ✔️ Correct.

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9. 2(x - 1)^(1/2) - 3 = 7

Add 3 to both sides:
2√(x - 1) = 10

Divide by 2:
√(x - 1) = 5

Square both sides:
x - 1 = 25 → x = 26

Check: 2√(26 - 1) - 3 = 2√25 - 3 = 2×5 - 3 = 10 - 3 = 7 → ✔️ Correct.

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10. ∛x + 1 = -2

Subtract 1:
∛x = -3

Cube both sides:
x = (-3)^3 = -27

Check: ∛(-27) + 1 = -3 + 1 = -2 → ✔️ Correct.

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11. 4∛x + 2 = 0

Subtract 2:
4∛x = -2

Divide by 4:
∛x = -1/2

Cube both sides:
x = (-1/2)^3 = -1/8

Check: 4∛(-1/8) + 2 = 4*(-1/2) + 2 = -2 + 2 = 0 → ✔️ Correct.

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12. ∛(2x + 7) = 5

Cube both sides:
2x + 7 = 125

Subtract 7:
2x = 118 → x = 59

Check: ∛(2*59 + 7) = ∛(118 + 7) = ∛125 = 5 → ✔️ Correct.

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13. (x + 4)^(1/3) - 2 = -6

Add 2:
(x + 4)^(1/3) = -4

Cube both sides:
x + 4 = (-4)^3 = -64 → x = -68

Check: ∛(-68 + 4) - 2 = ∛(-64) - 2 = -4 - 2 = -6 → ✔️ Correct.

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14. 8∛x + 3 = 11

Subtract 3:
8∛x = 8

Divide by 8:
∛x = 1

Cube both sides:
x = 1

Check: 8∛1 + 3 = 8*1 + 3 = 11 → ✔️ Correct.

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15. 3x^(1/3) - 2 = -4

Add 2:
3x^(1/3) = -2

Divide by 3:
x^(1/3) = -2/3

Cube both sides:
x = (-2/3)^3 = -8/27

Check: 3∛(-8/27) - 2 = 3*(-2/3) - 2 = -2 - 2 = -4 → ✔️ Correct.

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16. -2∛(2x + 5) + 7 = 15

Subtract 7:
-2∛(2x + 5) = 8

Divide by -2:
∛(2x + 5) = -4

Cube both sides:
2x + 5 = (-4)^3 = -64 → 2x = -69 → x = -69/2

Check: -2∛(2*(-69/2) + 5) + 7 = -2∛(-69 + 5) + 7 = -2∛(-64) + 7 = -2*(-4) + 7 = 8 + 7 = 15 → ✔️ Correct.

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17. (1/2)(5x + 1)^(1/3) + 5/2 = 4

Subtract 5/2:
(1/2)(5x + 1)^(1/3) = 4 - 5/2 = 3/2

Multiply both sides by 2:
(5x + 1)^(1/3) = 3

Cube both sides:
5x + 1 = 27 → 5x = 26 → x = 26/5

Check: (1/2)∛(5*(26/5) + 1) + 5/2 = (1/2)∛(26 + 1) + 5/2 = (1/2)*3 + 5/2 = 3/2 + 5/2 = 8/2 = 4 → ✔️ Correct.

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18. 6∛(x - 3) + 2 = 1/2

Subtract 2:
6∛(x - 3) = 1/2 - 2 = -3/2

Divide by 6:
∛(x - 3) = (-3/2)/6 = -3/12 = -1/4

Cube both sides:
x - 3 = (-1/4)^3 = -1/64 → x = 3 - 1/64 = 192/64 - 1/64 = 191/64

Check: 6∛(191/64 - 3) + 2 = 6∛(191/64 - 192/64) + 2 = 6∛(-1/64) + 2 = 6*(-1/4) + 2 = -6/4 + 2 = -3/2 + 2 = 1/2 → ✔️ Correct.

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Now let’s do the ones that say “Check for extraneous solutions” — these often involve even roots or squaring, which can introduce false answers.

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19. x^(5/3) = 243

This is same as (∛x)^5 = 243 OR ∛(x^5) = 243 — easier to write as:

Raise both sides to 3/5 power:
x = 243^(3/5)

Note: 243 = 3^5 → so 243^(3/5) = (3^5)^(3/5) = 3^3 = 27

Check: 27^(5/3) = (∛27)^5 = 3^5 = 243 → ✔️ Correct.

No extraneous solution here since odd root and odd power.

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20. x^(3/2) + 3 = 11

Subtract 3:
x^(3/2) = 8

Which is (√x)^3 = 8 → take cube root: √x = 2 → x = 4

OR raise both sides to 2/3: x = 8^(2/3) = (8^(1/3))^2 = 2^2 = 4

Check: 4^(3/2) = (√4)^3 = 2^3 = 8 → 8 + 3 = 11 → ✔️ Correct.

But note: x must be ≥ 0 because of square root in exponent. x=4 is fine.

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21. 2x^(5/3) = -64

Divide by 2:
x^(5/3) = -32

Raise both sides to 3/5:
x = (-32)^(3/5)

Note: -32 = (-2)^5 → so [(-2)^5]^(3/5) = (-2)^3 = -8

Check: 2*(-8)^(5/3) = 2*[∛(-8)]^5 = 2*(-2)^5 = 2*(-32) = -64 → ✔️ Correct.

Odd root allows negative base — no extraneous solution.

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22. (x - 2)^(3/4) = 8

This is [⁴√(x - 2)]^3 = 8

Take cube root first: ⁴√(x - 2) = 2

Then raise to 4th power: x - 2 = 16 → x = 18

Check: (18 - 2)^(3/4) = 16^(3/4) = (⁴√16)^3 = 2^3 = 8 → ✔️ Correct.

Also, domain: x - 2 ≥ 0 → x ≥ 2 → 18 is okay.

Is there another solution? Since it's an even root (4th), but raised to odd power (3), still only one real solution. No extraneous.

Wait — actually, when you have fractional exponents with even denominator, you must ensure the base is non-negative. Here, after solving, x=18 works. But could there be a negative base? No, because 4th root of negative is not real. So only x=18.

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23. (2x + 12)^(2/3) - 3 = 13

Add 3:
(2x + 12)^(2/3) = 16

This is [∛(2x + 12)]^2 = 16

So ∛(2x + 12) = ±4

Case 1: ∛(2x + 12) = 4 → 2x + 12 = 64 → 2x = 52 → x = 26

Case 2: ∛(2x + 12) = -4 → 2x + 12 = -64 → 2x = -76 → x = -38

Now check both:

For x=26: (2*26 + 12)^(2/3) = (52+12)^(2/3)=64^(2/3)=(∛64)^2=4^2=16 → 16-3=13 ✔️

For x=-38: (2*(-38)+12)^(2/3)=(-76+12)^(2/3)=(-64)^(2/3)=[∛(-64)]^2=(-4)^2=16 → 16-3=13 ✔️

Both work! Because even though inside cube root is negative, we’re squaring it afterward, so result is positive. And cube root of negative is defined.

So two solutions: x=26 and x=-38.

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24. (3x + 21)^(4/3) + 9 = 90

Subtract 9:
(3x + 21)^(4/3) = 81

This is [∛(3x + 21)]^4 = 81

Take 4th root: ∛(3x + 21) = ±3 (since 3^4=81, (-3)^4=81)

Case 1: ∛(3x + 21) = 3 → 3x + 21 = 27 → 3x = 6 → x = 2

Case 2: ∛(3x + 21) = -3 → 3x + 21 = -27 → 3x = -48 → x = -16

Check both:

x=2: (3*2 + 21)^(4/3) = (6+21)^(4/3)=27^(4/3)=(∛27)^4=3^4=81 → 81+9=90 ✔️

x=-16: (3*(-16)+21)^(4/3)=(-48+21)^(4/3)=(-27)^(4/3)=[∛(-27)]^4=(-3)^4=81 → 81+9=90 ✔️

Both work. Cube root handles negatives, then raising to 4th power makes positive.

Solutions: x=2, x=-16

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Now the last set — equations with radicals on both sides. These require squaring both sides, which can create extraneous solutions. Must check all answers.

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25. √(x - 3) = √(2x - 7)

Square both sides:
x - 3 = 2x - 7

Solve:
-3 + 7 = 2x - x → 4 = x

Check: Left: √(4-3)=√1=1; Right: √(8-7)=√1=1 → ✔️ Valid.

Only solution: x=4

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26. √(x + 3) = √(4x - 8)

Square both sides:
x + 3 = 4x - 8

Solve:
3 + 8 = 4x - x → 11 = 3x → x = 11/3

Check domains first:
Left: x + 3 ≥ 0 → x ≥ -3 → ok
Right: 4x - 8 ≥ 0 → x ≥ 2 → 11/3 ≈ 3.67 ≥ 2 → ok

Now plug in:
Left: √(11/3 + 3) = √(11/3 + 9/3) = √(20/3)
Right: √(4*(11/3) - 8) = √(44/3 - 24/3) = √(20/3) → equal ✔️

Solution: x = 11/3

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27. ∛(4x - 9) = ∛(2x - 4)

Since cube roots are one-to-one, we can just set insides equal:
4x - 9 = 2x - 4

Solve:
4x - 2x = -4 + 9 → 2x = 5 → x = 5/2

Check:
Left: ∛(4*(5/2) - 9) = ∛(10 - 9) = ∛1 = 1
Right: ∛(2*(5/2) - 4) = ∛(5 - 4) = ∛1 = 1 → ✔️

Solution: x = 5/2

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28. ∛(3x + 3) = ∛(2x - 7)

Set insides equal:
3x + 3 = 2x - 7

Solve:
3x - 2x = -7 - 3 → x = -10

Check:
Left: ∛(3*(-10) + 3) = ∛(-30 + 3) = ∛(-27) = -3
Right: ∛(2*(-10) - 7) = ∛(-20 - 7) = ∛(-27) = -3 → ✔️

Solution: x = -10

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29. √x + 1 = √(3x - 3)

Isolate one radical? Actually, let’s move terms:

√x - √(3x - 3) = -1 → maybe better to square both sides directly.

Original: √x + 1 = √(3x - 3)

Square both sides:
(√x + 1)^2 = (√(3x - 3))^2
→ x + 2√x + 1 = 3x - 3

Bring all to one side:
x + 2√x + 1 - 3x + 3 = 0 → -2x + 2√x + 4 = 0

Divide by 2:
-x + √x + 2 = 0 → √x = x - 2

Now square again:
x = (x - 2)^2 = x^2 - 4x + 4

Bring all to one side:
0 = x^2 - 5x + 4 → factor: (x - 1)(x - 4) = 0 → x=1 or x=4

Now CHECK BOTH (because we squared twice):

x=1: Left: √1 + 1 = 1 + 1 = 2; Right: √(3*1 - 3) = √0 = 0 → 2 ≠ 0 ✘ Extraneous

x=4: Left: √4 + 1 = 2 + 1 = 3; Right: √(12 - 3) = √9 = 3 → ✔️ Valid

Only solution: x=4

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30. ∛(3x + 9) = ∛(x + 6)

Set insides equal:
3x + 9 = x + 6 → 2x = -3 → x = -3/2

Check:
Left: ∛(3*(-3/2) + 9) = ∛(-9/2 + 18/2) = ∛(9/2)
Right: ∛(-3/2 + 6) = ∛(-3/2 + 12/2) = ∛(9/2) → equal ✔️

Solution: x = -3/2

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31. x + 2 = √(2x + 7)

Square both sides:
(x + 2)^2 = 2x + 7
→ x^2 + 4x + 4 = 2x + 7

Bring all to left:
x^2 + 2x - 3 = 0 → factor: (x + 3)(x - 1) = 0 → x = -3 or x = 1

Check:

x = -3: Left: -3 + 2 = -1; Right: √(2*(-3) + 7) = √(1) = 1 → -1 ≠ 1 ✘ Extraneous

x = 1: Left: 1 + 2 = 3; Right: √(2 + 7) = √9 = 3 → ✔️ Valid

Only solution: x = 1

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32. √(2x + 3) = 1 + √(x + 1)

Square both sides:
[√(2x + 3)]^2 = [1 + √(x + 1)]^2
→ 2x + 3 = 1 + 2√(x + 1) + (x + 1)
→ 2x + 3 = x + 2 + 2√(x + 1)

Simplify:
2x + 3 - x - 2 = 2√(x + 1) → x + 1 = 2√(x + 1)

Let u = √(x + 1), then u^2 = x + 1 → equation becomes:
u^2 = 2u → u^2 - 2u = 0 → u(u - 2) = 0 → u=0 or u=2

So:

If u=0 → √(x+1)=0 → x+1=0 → x=-1

If u=2 → √(x+1)=2 → x+1=4 → x=3

Check both:

x=-1: Left: √(2*(-1)+3)=√(1)=1; Right: 1 + √(-1+1)=1+0=1 → ✔️ Valid

x=3: Left: √(6+3)=√9=3; Right: 1 + √(3+1)=1+2=3 → ✔️ Valid

Both work!

Solutions: x = -1, x = 3

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Final Answer:
Here are all the solutions for problems 1 through 32:

1. x = 81
2. x = 25
3. x = 2
4. x = 3/2
5. x = 3
6. x = -15
7. x = 173
8. x = -7/12
9. x = 26
10. x = -27
11. x = -1/8
12. x = 59
13. x = -68
14. x = 1
15. x = -8/27
16. x = -69/2
17. x = 26/5
18. x = 191/64
19. x = 27
20. x = 4
21. x = -8
22. x = 18
23. x = 26, x = -38
24. x = 2, x = -16
25. x = 4
26. x = 11/3
27. x = 5/2
28. x = -10
29. x = 4
30. x = -3/2
31. x = 1
32. x = -1, x = 3