Solving Radical Equations

Worked solutions · Section 4.4, page 222 · Problems 1-36

Solving a radical equation

\(n\) = index (the small number on the radical)

1. Radical equation

\(2\sqrt{x+1} = 4\)
Variable under the radical. A radical holding only numbers is a coefficient

2. Isolate the radical

\(\sqrt[n]{u} + c = d \ \Rightarrow\ \sqrt[n]{u} = d - c\)
Radical alone on one side before any power

3. Raise each side to the index

\(\left(\sqrt[n]{u}\right)^n = u\)
Every side, not only the radical
  • \(\sqrt[3]{u} = 4 \Rightarrow u = 4^3 = 64\)
  • Index 2: square. Index 3: cube. Index 4: fourth power

4. Square of a binomial

\((a+b)^2 = a^2 + 2ab + b^2\)
A radical left after squaring: isolate it and square again

5. Two radicals, same index

\(\sqrt[n]{u} = \sqrt[n]{v} \ \Rightarrow\ u = v\)
Equate the radicands

6. Zero-product property

\(ab = 0 \ \Rightarrow\ a = 0 \ \text{or}\ b = 0\)
Standard form first, then factor

Rational exponent equations

7. Rational exponent = radical

\(x^{m/n} = \sqrt[n]{x^m}\)
Denominator = index

8. Raise to the reciprocal exponent

\(\left(x^{m/n}\right)^{n/m} = x\)
Isolate the power first; a coefficient is not part of it

9. Even numerator: two solutions

\(x^{2/3} = c \ \Rightarrow\ x = \pm\, c^{3/2}\)
\(\left(x^{1/3}\right)^2 = c\) keeps both signs

Checking and sign rules

10. Check every apparent solution

\(\sqrt{5x+1} = 6 \ \Rightarrow\ \sqrt{5(7)+1} = \sqrt{36} = 6\)
Raising to a power can add false solutions

11. Extraneous solution

\(x + 1 = \sqrt{7x+15} \ \Rightarrow\ x = 7 \ \text{only},\ x = -2 \ \text{fails}\)
Apparent solution that fails the original equation. Discard it

12. Even index is never negative

\(\sqrt[n]{u} \ge 0 \quad (n \text{ even})\)
Set equal to a negative number: no real solution

13. Odd index keeps the sign

\(\sqrt[n]{u} = c \quad (n \text{ odd})\)
Any sign allowed; cubing adds no extraneous solution

14. Radicand of an even root

\(u \ge 0 \quad (n \text{ even})\)
An inequality needs this too: under an even radical, \(u\) cannot be negative

Problems 1-2: Vocabulary and Core Concept Check

Problem 1

\(3x - \sqrt{2} = \sqrt{6}\)

Rule used:
  • 1. Radical equation
Step 1. Look under each radical.
\(\sqrt{2},\quad \sqrt{6}\)
Step 2. No variable in either radicand.
\(x\) sits outside both radicals
Answer
\(Not a radical equation\)
Not a radical equation; no variable under a radical
Problem 2

\(\sqrt{x} + 10 < 15\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 14. Radicand of an even root
Step 1. Isolate the radical.
\(\sqrt{x} < 5\)
Step 2. Square each side.
\(x < 25\)
Step 3. Keep the radicand nonnegative.
\(x \ge 0\)
Answer
\(0 \le x < 25\)

Problems 3-12: Solve the equation. Check your solution.

Problem 3

\(\sqrt{5x+1} = 6\)

Rules used:
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Square each side.
\(\left(\sqrt{5x+1}\right)^2 = 6^2\)
\(5x + 1 = 36\)
Step 2. Solve for x.
\(5x = 35\)
\(x = 7\)
Step 3. Check.
\(\sqrt{5(7)+1} = \sqrt{36} = 6\)
Solution
\(x = 7\)
Problem 4

\(\sqrt{3x+10} = 8\)

Rules used:
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Square each side.
\(3x + 10 = 64\)
Step 2. Solve for x.
\(3x = 54\)
\(x = 18\)
Step 3. Check.
\(\sqrt{3(18)+10} = \sqrt{64} = 8\)
Solution
\(x = 18\)
Problem 5

\(\sqrt[3]{x-16} = 2\)

Rules used:
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Cube each side.
\(x - 16 = 2^3\)
\(x - 16 = 8\)
Step 2. Solve for x.
\(x = 24\)
Step 3. Check.
\(\sqrt[3]{24-16} = \sqrt[3]{8} = 2\)
Solution
\(x = 24\)
Problem 6

\(\sqrt[3]{x} - 10 = -7\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Isolate the radical.
\(\sqrt[3]{x} = 3\)
Step 2. Cube each side.
\(x = 3^3\)
\(x = 27\)
Step 3. Check.
\(\sqrt[3]{27} - 10 = 3 - 10 = -7\)
Solution
\(x = 27\)
Problem 7

\(-2\sqrt{24x} + 13 = -11\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Isolate the radical.
\(-2\sqrt{24x} = -24\)
\(\sqrt{24x} = 12\)
Step 2. Square each side.
\(24x = 144\)
Step 3. Solve for x.
\(x = 6\)
Step 4. Check.
\(-2\sqrt{24(6)} + 13 = -2(12) + 13 = -11\)
Solution
\(x = 6\)
Problem 8

\(8\sqrt[3]{10x} - 15 = 17\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Isolate the radical.
\(8\sqrt[3]{10x} = 32\)
\(\sqrt[3]{10x} = 4\)
Step 2. Cube each side.
\(10x = 4^3\)
\(10x = 64\)
Step 3. Solve for x.
\(x = \dfrac{32}{5}\)
Step 4. Check.
\(8\sqrt[3]{10\left(\frac{32}{5}\right)} - 15 = 8(4) - 15 = 17\)
Solution
\(x = \dfrac{32}{5} = 6.4\)
Problem 9

\(\dfrac{1}{5}\sqrt[3]{3x} + 10 = 8\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 13. Odd index keeps the sign
Step 1. Isolate the radical.
\(\dfrac{1}{5}\sqrt[3]{3x} = -2\)
\(\sqrt[3]{3x} = -10\)
Step 2. Cube each side.
\(3x = (-10)^3\)
\(3x = -1000\)
Step 3. Solve for x.
\(x = -\dfrac{1000}{3}\)
Step 4. Check.
\(\dfrac{1}{5}\sqrt[3]{3\left(-\frac{1000}{3}\right)} + 10 = \dfrac{-10}{5} + 10 = 8\)
Solution
\(x = -\dfrac{1000}{3}\)
Problem 10

\(\sqrt{2x} - \dfrac{2}{3} = 0\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Isolate the radical.
\(\sqrt{2x} = \dfrac{2}{3}\)
Step 2. Square each side.
\(2x = \dfrac{4}{9}\)
Step 3. Solve for x.
\(x = \dfrac{2}{9}\)
Step 4. Check.
\(\sqrt{2\left(\frac{2}{9}\right)} = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}\)
Solution
\(x = \dfrac{2}{9}\)
Problem 11

\(2\sqrt[5]{x} + 7 = 15\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Isolate the radical.
\(2\sqrt[5]{x} = 8\)
\(\sqrt[5]{x} = 4\)
Step 2. Raise each side to the fifth power.
\(x = 4^5\)
\(x = 1024\)
Step 3. Check.
\(2\sqrt[5]{1024} + 7 = 2(4) + 7 = 15\)
Solution
\(x = 1024\)
Problem 12

\(\sqrt[4]{4x} - 13 = -15\)

Rules used:
  • 2. Isolate the radical
  • 12. Even index is never negative
Step 1. Isolate the radical.
\(\sqrt[4]{4x} = -2\)
Step 2. Read the index.
Index 4 is even, so the root is never negative
Step 3. Reject the apparent solution.
\(4x = (-2)^4 = 16\)
\(x = 4\)
\(\sqrt[4]{4(4)} - 13 = 2 - 13 = -11\), not \(-15\)
Solution
\(\text{no real solution}\)

Problems 13-14: Modeling with mathematics

Problem 13

\(h = 62.5\sqrt[3]{t} + 75.8\)

Rules used:
  • 2. Isolate the radical
  • 3. Raise each side to the index
  • 13. Odd index keeps the sign
Step 1. Substitute 250 for h.
\(250 = 62.5\sqrt[3]{t} + 75.8\)
Step 2. Isolate the radical.
\(174.2 = 62.5\sqrt[3]{t}\)
\(\sqrt[3]{t} = 2.7872\)
Step 3. Cube each side.
\(t = 2.7872^3 \approx 21.7\)
Answer
\(t \approx 22\)
About 22 years old
Problem 14

\(v = \sqrt{2gh}\)

Rules used:
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Substitute 15 for v and 9.8 for g.
\(15 = \sqrt{2(9.8)h}\)
\(15 = \sqrt{19.6h}\)
Step 2. Square each side.
\(225 = 19.6h\)
Step 3. Divide each side by 19.6.
\(h = \dfrac{225}{19.6} \approx 11.5\)
Answer
\(h \approx 11.5\)
About 11.5 meters

Problems 15-26: Solve the equation. Check your solution(s).

Squaring can add a solution the original equation does not have. Check each one

Problem 15

\(x - 6 = \sqrt{3x}\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\((x-6)^2 = 3x\)
\(x^2 - 12x + 36 = 3x\)
Step 2. Write in standard form.
\(x^2 - 15x + 36 = 0\)
Step 3. Factor.
\((x-12)(x-3) = 0\)
\(x = 12\)
Step 4. Check both apparent solutions.
\(12 - 6 = \sqrt{36} = 6\)
\(3 - 6 = -3\), and \(\sqrt{9} = 3\), so 3 is extraneous
Solution
\(x = 12\)
Problem 16

\(x - 10 = \sqrt{9x}\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\((x-10)^2 = 9x\)
\(x^2 - 20x + 100 = 9x\)
Step 2. Write in standard form.
\(x^2 - 29x + 100 = 0\)
Step 3. Factor.
\((x-25)(x-4) = 0\)
\(x = 25\)
Step 4. Check both apparent solutions.
\(25 - 10 = \sqrt{225} = 15\)
\(4 - 10 = -6\), and \(\sqrt{36} = 6\), so 4 is extraneous
Solution
\(x = 25\)
Problem 17

\(\sqrt{44-2x} = x - 10\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\(44 - 2x = (x-10)^2\)
\(44 - 2x = x^2 - 20x + 100\)
Step 2. Write in standard form.
\(0 = x^2 - 18x + 56\)
Step 3. Factor.
\(0 = (x-14)(x-4)\)
\(x = 14\)
Step 4. Check both apparent solutions.
\(\sqrt{44-2(14)} = \sqrt{16} = 4 = 14 - 10\)
\(\sqrt{36} = 6\), and \(4 - 10 = -6\), so 4 is extraneous
Solution
\(x = 14\)
Problem 18

\(\sqrt{2x+30} = x + 3\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\(2x + 30 = (x+3)^2\)
\(2x + 30 = x^2 + 6x + 9\)
Step 2. Write in standard form.
\(0 = x^2 + 4x - 21\)
Step 3. Factor.
\(0 = (x+7)(x-3)\)
\(x = 3\)
Step 4. Check both apparent solutions.
\(\sqrt{2(3)+30} = \sqrt{36} = 6 = 3 + 3\)
\(\sqrt{16} = 4\), and \(-7 + 3 = -4\), so \(-7\) is extraneous
Solution
\(x = 3\)
Problem 19

\(\sqrt[3]{8x^3-1} = 2x - 1\)

Rules used:
  • 3. Raise each side to the index
  • 6. Zero-product property
  • 13. Odd index keeps the sign
Step 1. Cube each side.
\(8x^3 - 1 = (2x-1)^3\)
\(8x^3 - 1 = 8x^3 - 12x^2 + 6x - 1\)
Step 2. Write in standard form.
\(0 = -12x^2 + 6x\)
\(0 = (6x)(1 - 2x)\)
Step 3. Apply the zero-product property.
\(x = 0\)
or \(1 - 2x = 0\)
Step 4. Check.
\(\sqrt[3]{-1} = -1 = 2(0) - 1\)
\(x = \frac{1}{2}\): \(\sqrt[3]{0} = 0 = 2\left(\frac{1}{2}\right) - 1\)
Solutions
\(x = 0 \ \text{or}\ x = \dfrac{1}{2}\)
Problem 20

\(\sqrt[4]{3-8x^2} = 2x\)

Rules used:
  • 3. Raise each side to the index
  • 6. Zero-product property
  • 12. Even index is never negative
Step 1. Raise each side to the fourth power.
\(3 - 8x^2 = (2x)^4\)
\(3 - 8x^2 = 16x^4\)
Step 2. Write in standard form.
\(16x^4 + 8x^2 - 3 = 0\)
Step 3. Factor.
\((4x^2 - 1)(4x^2 + 3) = 0\)
\(4x^2 = 1\)
Step 4. Keep the root that makes 2x nonnegative.
\(x = \dfrac{1}{2}\)
\(x = -\frac{1}{2}\) gives \(2x = -1\), and a fourth root is never negative
Solution
\(x = \dfrac{1}{2}\)
Problem 21

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

Rules used:
  • 5. Two radicals, same index
  • 10. Check every apparent solution
Step 1. Equate the radicands.
\(4x + 1 = x + 10\)
Step 2. Solve for x.
\(3x = 9\)
\(x = 3\)
Step 3. Check.
\(\sqrt{4(3)+1} = \sqrt{13} = \sqrt{3+10}\)
Solution
\(x = 3\)
Problem 22

\(\sqrt{3x-3} - \sqrt{x+12} = 0\)

Rules used:
  • 2. Isolate the radical
  • 5. Two radicals, same index
  • 10. Check every apparent solution
Step 1. Isolate one radical.
\(\sqrt{3x-3} = \sqrt{x+12}\)
Step 2. Equate the radicands.
\(3x - 3 = x + 12\)
Step 3. Solve for x.
\(2x = 15\)
\(x = \dfrac{15}{2}\)
Step 4. Check.
\(3\left(\frac{15}{2}\right) - 3 = \dfrac{39}{2} = \dfrac{15}{2} + 12\)
Solution
\(x = \dfrac{15}{2} = 7.5\)
Problem 23

\(\sqrt[3]{2x-5} - \sqrt[3]{8x+1} = 0\)

Rules used:
  • 2. Isolate the radical
  • 5. Two radicals, same index
  • 13. Odd index keeps the sign
Step 1. Isolate one radical.
\(\sqrt[3]{2x-5} = \sqrt[3]{8x+1}\)
Step 2. Equate the radicands.
\(2x - 5 = 8x + 1\)
Step 3. Solve for x.
\(-6 = 6x\)
\(x = -1\)
Step 4. Check.
\(2(-1) - 5 = -7 = 8(-1) + 1\)
Solution
\(x = -1\)
Problem 24

\(\sqrt[3]{x+5} = 2\sqrt[3]{2x+6}\)

Rules used:
  • 3. Raise each side to the index
  • 13. Odd index keeps the sign
Step 1. Cube each side.
\(x + 5 = 2^3(2x+6)\)
\(x + 5 = 16x + 48\)
Step 2. Solve for x.
\(-43 = 15x\)
\(x = -\dfrac{43}{15}\)
Step 3. Check.
\(\sqrt[3]{\dfrac{32}{15}} = 2\sqrt[3]{\dfrac{4}{15}}\)
Solution
\(x = -\dfrac{43}{15}\)
Problem 25

\(\sqrt{3x-8} + 1 = \sqrt{x+5}\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\(3x - 8 + 2\sqrt{3x-8} + 1 = x + 5\)
Step 2. Isolate the remaining radical.
\(2\sqrt{3x-8} = -2x + 12\)
\(\sqrt{3x-8} = -x + 6\)
Step 3. Square each side again.
\(3x - 8 = x^2 - 12x + 36\)
Step 4. Write in standard form and factor.
\(0 = x^2 - 15x + 44\)
\(0 = (x-11)(x-4)\)
\(x = 4\)
Step 5. Check both apparent solutions.
\(\sqrt{3(4)-8} + 1 = 2 + 1 = 3 = \sqrt{4+5}\)
\(x = 11\): \(\sqrt{25} + 1 = 6\), and \(\sqrt{16} = 4\), so 11 is extraneous
Solution
\(x = 4\)
Problem 26

\(\sqrt{x+2} = 2 - \sqrt{x}\)

Rules used:
  • 3. Raise each side to the index
  • 4. Square of a binomial
  • 10. Check every apparent solution
Step 1. Square each side.
\(x + 2 = 4 - 4\sqrt{x} + x\)
Step 2. Isolate the remaining radical.
\(4\sqrt{x} = 2\)
\(\sqrt{x} = \dfrac{1}{2}\)
Step 3. Square each side again.
\(x = \dfrac{1}{4}\)
Step 4. Check.
\(\sqrt{\dfrac{1}{4} + 2} = \dfrac{3}{2} = 2 - \dfrac{1}{2}\)
Solution
\(x = \dfrac{1}{4}\)

Problems 27-34: Solve the equation. Check your solution(s).

Rational exponents

Problem 27

\(2x^{2/3} = 8\)

Rules used:
  • 8. Raise to the reciprocal exponent
  • 9. Even numerator: two solutions
Step 1. Divide each side by 2.
\(x^{2/3} = 4\)
Step 2. Raise each side to the three-halves.
\(x = \pm\, 4^{3/2}\)
\(x = \pm\, 8\)
Step 3. Check.
\(2(8)^{2/3} = 2(4) = 8\)
\(2(-8)^{2/3} = 2(4) = 8\)
Solutions
\(x = 8 \ \text{or}\ x = -8\)
Problem 28

\(4x^{3/2} = 32\)

Rules used:
  • 2. Isolate the radical
  • 8. Raise to the reciprocal exponent
Step 1. Divide each side by 4.
\(x^{3/2} = 8\)
Step 2. Raise each side to the two-thirds.
\(x = 8^{2/3}\)
\(x = 4\)
Step 3. Check.
\(4(4)^{3/2} = 4(8) = 32\)
Solution
\(x = 4\)
Problem 29

\(x^{1/4} + 3 = 0\)

Rules used:
  • 2. Isolate the radical
  • 7. Rational exponent = radical
  • 12. Even index is never negative
Step 1. Isolate the power.
\(x^{1/4} = -3\)
Step 2. Read the exponent.
\(x^{1/4} = \sqrt[4]{x}\), and a fourth root is never negative
Step 3. Reject the apparent solution.
\(x = (-3)^4 = 81\)
\(81^{1/4} + 3 = 3 + 3 = 6\), not 0
Solution
\(\text{no real solution}\)
Problem 30

\(2x^{3/4} - 14 = 40\)

Rules used:
  • 2. Isolate the radical
  • 8. Raise to the reciprocal exponent
Step 1. Isolate the power.
\(2x^{3/4} = 54\)
\(x^{3/4} = 27\)
Step 2. Raise each side to the four-thirds.
\(x = 27^{4/3}\)
\(x = 81\)
Step 3. Check.
\(2(81)^{3/4} - 14 = 2(27) - 14 = 40\)
Solution
\(x = 81\)
Problem 31

\((x+6)^{1/2} = x\)

Rules used:
  • 8. Raise to the reciprocal exponent
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\(x + 6 = x^2\)
Step 2. Write in standard form and factor.
\(0 = x^2 - x - 6\)
\(0 = (x-3)(x+2)\)
\(x = 3\)
Step 3. Check both apparent solutions.
\((3+6)^{1/2} = 3\)
\(x = -2\): \((-2+6)^{1/2} = 2\), not \(-2\)
Solution
\(x = 3\)
Problem 32

\((5-x)^{1/2} - 2x = 0\)

Rules used:
  • 2. Isolate the radical
  • 8. Raise to the reciprocal exponent
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Isolate the power.
\((5-x)^{1/2} = 2x\)
Step 2. Square each side.
\(5 - x = 4x^2\)
Step 3. Write in standard form and factor.
\(4x^2 + x - 5 = 0\)
\((4x+5)(x-1) = 0\)
\(x = 1\)
Step 4. Check both apparent solutions.
\((5-1)^{1/2} - 2(1) = 2 - 2 = 0\)
\(x = -\frac{5}{4}\) gives \(2x < 0\), so it is extraneous
Solution
\(x = 1\)
Problem 33

\(2(x+11)^{1/2} = x + 3\)

Rules used:
  • 8. Raise to the reciprocal exponent
  • 4. Square of a binomial
  • 6. Zero-product property
  • 11. Extraneous solution
Step 1. Square each side.
\(4(x+11) = (x+3)^2\)
\(4x + 44 = x^2 + 6x + 9\)
Step 2. Write in standard form and factor.
\(0 = x^2 + 2x - 35\)
\(0 = (x+7)(x-5)\)
\(x = 5\)
Step 3. Check both apparent solutions.
\(2(5+11)^{1/2} = 2(4) = 8 = 5 + 3\)
\(x = -7\): \(2(4)^{1/2} = 4\), and \(-7 + 3 = -4\), so \(-7\) is extraneous
Solution
\(x = 5\)
Problem 34

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

Rules used:
  • 8. Raise to the reciprocal exponent
  • 6. Zero-product property
  • 12. Even index is never negative
Step 1. Raise each side to the fourth power.
\(5x^2 - 4 = x^4\)
Step 2. Write in standard form and factor.
\(x^4 - 5x^2 + 4 = 0\)
\((x^2-1)(x^2-4) = 0\)
Step 3. Keep the roots that make x nonnegative.
\(x = 1\)
\(x = 2\)
\(x = -1\) and \(x = -2\) fail: a fourth root is never negative
Step 4. Check.
\((5(1)^2-4)^{1/4} = 1\)
\((5(2)^2-4)^{1/4} = 16^{1/4} = 2\)
Solutions
\(x = 1 \ \text{or}\ x = 2\)

Problems 35-36: Describe and correct the error in solving the equation.

Problem 35 · Error analysis

\(\sqrt[3]{3x-8} = 4\)

The book's work
\(\sqrt[3]{3x-8} = 4\)
\(\left(\sqrt[3]{3x-8}\right)^3 = 4\)
\(3x - 8 = 4\)
\(3x = 12\)
\(x = 4\)
Rules used:
  • 3. Raise each side to the index
  • 10. Check every apparent solution
Step 1. Find the error.
Only the left side was cubed
Step 2. Cube each side.
\(3x - 8 = 4^3\)
\(3x - 8 = 64\)
Step 3. Solve for x.
\(3x = 72\)
\(x = 24\)
Step 4. Check.
\(\sqrt[3]{3(24)-8} = \sqrt[3]{64} = 4\)
Solution
\(x = 24\)
Problem 36 · Error analysis

\(8x^{3/2} = 1000\)

The book's work
\(8x^{3/2} = 1000\)
\(8\left(x^{3/2}\right)^{2/3} = 1000^{2/3}\)
\(8x = 100\)
\(x = \dfrac{25}{2}\)
Rules used:
  • 2. Isolate the radical
  • 8. Raise to the reciprocal exponent
Step 1. Find the error.
The coefficient 8 was left outside the power
Step 2. Divide each side by 8 first.
\(x^{3/2} = 125\)
Step 3. Raise each side to the two-thirds.
\(x = 125^{2/3}\)
\(x = 25\)
Step 4. Check.
\(8(25)^{3/2} = 8(125) = 1000\)
Solution
\(x = 25\)

Quick checklist

  1. Radical alone on one side? Isolate it before any power
  2. Index \(n\)? Raise every side to the \(n\)
  3. Radical left after squaring? Isolate it and square again
  4. Two radicals, same index? Equate the radicands
  5. Rational exponent \(m/n\)? Raise each side to \(n/m\), coefficient first
  6. Even index equal to a negative number? No real solution
  7. Quadratic? Standard form, factor, zero-product property
  8. Every apparent solution checked in the original equation? Extraneous ones go out