Rearranging Equations and Formulas

4. Rearranging Formulas with Powers and Roots

Introduction

Many mathematical and scientific formulas contain variables raised to powers or placed inside roots. Rearranging these formulas requires more than ordinary addition, subtraction, multiplication, and division. We must also understand how powers and roots act as inverse operations.

For example, consider the formula for the area of a circle:

A = πr²

This formula calculates the area when the radius is known. However, if we know the area and want to calculate the radius, we must rearrange the formula:

r² = A/π

r = √(A/π)

This process is called isolating a variable. It is an essential skill in algebra, geometry, physics, chemistry, engineering, and many other fields.

In this topic, we will explore how to rearrange formulas involving squares, cubes, higher powers, and roots. We will also learn why some rearrangements produce multiple solutions, how extraneous solutions can arise, and how to check that our answers are mathematically valid.

Understanding Powers and Roots

A power represents repeated multiplication of a number or variable by itself.

For example:

• x² = x × x

• x³ = x × x × x

• x⁴ = x × x × x × x

The small raised number is called the exponent or power.

In the expression x³:

• x is the base.

• 3 is the exponent.

• x³ means x multiplied by itself three times.

A root is an operation that reverses raising a number to a power, with some important restrictions.

Examples:

• √25 = 5

• ∛27 = 3

• ⁴√16 = 2

The square root symbol √ represents the principal square root, which is always non-negative for real numbers.

This distinction is important:

√25 = 5

But the equation x² = 25 has two solutions:

x = 5 or x = −5

Both numbers produce 25 when squared.

Powers and Roots as Inverse Operations

Inverse operations undo one another.

For example:

• Addition is reversed by subtraction.

• Multiplication is reversed by division.

• Squaring is reversed by taking square roots, with attention to signs.

• Cubing is reversed by taking cube roots.

Consider:

x³ = 64

To isolate x, take the cube root of both sides:

x = ∛64

x = 4

We can verify:

4³ = 64

The solution is correct.

Now consider:

x² = 64

Taking square roots gives:

x = ±√64

x = ±8

Both solutions satisfy the original equation:

8² = 64

(−8)² = 64

The ± symbol means positive or negative.

Important distinction between even and odd powers

For real numbers:

• Even powers can produce two real solutions, one positive and one negative.

• Odd powers produce exactly one real solution.

• An even power cannot equal a negative number.

Examples:

x² = 49 → x = ±7

x³ = 125 → x = 5

x⁴ = 81 → x = ±3

x² = −9 → no real solution

A negative number can be cubed to produce a negative result:

(−3)³ = −27

But squaring a real number always produces a non-negative result.

Isolating Squared Variables

When a variable is squared, the objective is to remove the exponent while keeping the equation balanced.

The general procedure is:

1. Move all other terms away from the squared variable.

2. Divide by any coefficient multiplying the squared expression.

3. Take the square root of both sides.

4. Include both positive and negative solutions when appropriate.

5. Check the solutions in the original equation.

Worked Example: Simple squared variable

Solve:

x² = 144

Take the square root of both sides:

x = ±√144

x = ±12

Answer: x = 12 or x = −12.

Worked Example: Squared variable with a coefficient

Solve:

5x² = 180

Divide both sides by 5:

x² = 36

Take square roots:

x = ±√36

x = ±6

Answer: x = ±6.

Verification:

5(6²) = 180

5(−6)² = 180

Both are correct.

Worked Example: Squared variable with addition

Solve:

3x² + 7 = 82

Subtract 7:

3x² = 75

Divide by 3:

x² = 25

Take square roots:

x = ±5

Answer: x = 5 or x = −5.

Worked Example: Squared expression involving brackets

Solve:

(x − 4)² = 49

Take square roots:

x − 4 = ±7

Solve the two possibilities separately.

Positive case:

x − 4 = 7

x = 11

Negative case:

x − 4 = −7

x = −3

Answer: x = 11 or x = −3.

Notice that the square applies to the entire expression (x − 4), not just to x.

Rearranging Geometric Formulas with Squares

Many geometric formulas involve squared lengths.

Worked Example: Finding the radius of a circle

The area of a circle is:

A = πr²

Rearrange to make r the subject.

Divide by π:

A/π = r²

Take square roots:

r = ±√(A/π)

Because radius is a physical length, it cannot be negative.

Therefore:

r = √(A/π)

Suppose a circular garden has an area of 78.5 m².

Using π ≈ 3.14:

r = √(78.5/3.14)

r = √25

r = 5 m

The radius is 5 metres.

Worked Example: Rearranging Pythagoras' theorem

For a right-angled triangle:

c² = a² + b²

Suppose we want to make a the subject.

Subtract b²:

a² = c² − b²

Take square roots:

a = √(c² − b²)

We use the positive root because a represents a side length.

If c = 13 cm and b = 5 cm:

a = √(13² − 5²)

a = √(169 − 25)

a = √144

a = 12 cm

The missing side is 12 cm.

Rearranging Formulas with Higher Powers

Not all formulas contain squares. Some involve cubes, fourth powers, or other exponents.

The general rule is:

If xⁿ = a, then x is found using the nth root of a, with the possible real solutions determined by whether n is even or odd.

Worked Example: Cubed variable

Solve:

2x³ = 250

Divide by 2:

x³ = 125

Take cube roots:

x = ∛125

x = 5

Verification:

2(5³) = 250

Worked Example: Fourth power

Solve:

3x⁴ = 243

Divide by 3:

x⁴ = 81

Take fourth roots:

x = ±⁴√81

x = ±3

Verification:

3(3⁴) = 243

3(−3)⁴ = 243

Both solutions are valid.

Worked Example: Rearranging a volume formula

The volume of a cube is:

V = s³

where s is the side length.

Make s the subject:

s = ∛V

Suppose the cube has a volume of 216 cm³.

s = ∛216

s = 6 cm

Only the positive answer is physically meaningful because a length cannot be negative.

Rearranging Formulas Involving Fractional Exponents

Roots can also be written as fractional powers.

For non-negative x:

√x = x^(1/2)

∛x = x^(1/3)

⁴√x = x^(1/4)

More generally, for suitable real-number domains:

x^(m/n) = (ⁿ√x)ᵐ

For example:

x^(3/2) = (√x)³

Fractional exponents are especially useful in advanced algebra and scientific formulas.

Worked Example: Fractional exponent

Solve:

x^(3/2) = 27

Raise both sides to the power 2/3:

x = 27^(2/3)

Calculate the cube root first:

∛27 = 3

Then square:

x = 3²

x = 9

Verification:

9^(3/2) = (√9)³ = 3³ = 27

Answer: x = 9.

Worked Example: Variable raised to 2/3

Solve:

x^(2/3) = 16

For real x, this means:

(∛x)² = 16

Take square roots:

∛x = ±4

Cube both sides:

x = 64 or x = −64

Both values satisfy the original equation.

This example shows why fractional exponents must be handled carefully: raising both sides to a reciprocal power without considering signs can lose solutions.

Rearranging Formulas Containing Square Roots

Sometimes the variable we want to isolate appears inside a square root.

The key technique is to isolate the square root first, then square both sides.

For example:

y = √(x + 5)

Square both sides:

y² = x + 5

Subtract 5:

x = y² − 5

Because the original square root is non-negative, this rearrangement requires y ≥ 0.

Worked Example: Square root with a coefficient

Rearrange to make x the subject:

y = 3√(x − 2)

Divide by 3:

y/3 = √(x − 2)

Square both sides:

y²/9 = x − 2

Add 2:

x = y²/9 + 2

This formula applies when y ≥ 0, and it automatically gives x ≥ 2.

Worked Example: A multi-step square root equation

Solve:

2√(x + 4) + 3 = 13

Subtract 3:

2√(x + 4) = 10

Divide by 2:

√(x + 4) = 5

Square both sides:

x + 4 = 25

Subtract 4:

x = 21

Verification:

2√(21 + 4) + 3 = 2(5) + 3 = 13

Answer: x = 21.

Extraneous Solutions

An extraneous solution is a value obtained during algebraic manipulation that does not satisfy the original equation.

Extraneous solutions often arise when we square both sides of an equation.

This happens because squaring removes information about whether a number was positive or negative.

For example:

3² = 9

(−3)² = 9

Although 3 and −3 are different, squaring produces the same result.

Worked Example: Identifying an extraneous solution

Solve:

√(x + 5) = x − 1

Square both sides:

x + 5 = (x − 1)²

Expand:

x + 5 = x² − 2x + 1

Rearrange:

0 = x² − 3x − 4

Factor:

0 = (x − 4)(x + 1)

Possible solutions:

x = 4 or x = −1

Now check both in the original equation.

For x = 4:

√(4 + 5) = 4 − 1

3 = 3 ✓

For x = −1:

√(−1 + 5) = −1 − 1

2 = −2 ✗

Therefore, x = −1 is extraneous.

Final answer: x = 4.

The original equation requires x − 1 ≥ 0, so x must be at least 1. This restriction could also have helped us reject −1.

Why checking is essential

Squaring both sides is a valid consequence of an equation, but the reverse implication does not always hold.

If a = b, then a² = b².

However, a² = b² only guarantees that a = b or a = −b.

Therefore, every solution obtained after squaring should be checked in the original equation.

Real-World Applications

Physics: Finding speed from kinetic energy

The kinetic energy of a moving object is:

KE = ½mv²

where:

• KE = kinetic energy (J)

• m = mass (kg)

• v = speed (m/s)

Make v the subject.

Multiply by 2:

2KE = mv²

Divide by m:

2KE/m = v²

Take square roots:

v = √(2KE/m)

We use the positive root because speed is non-negative.

Suppose a 4 kg object has 200 J of kinetic energy.

v = √(2 × 200/4)

v = √100

v = 10 m/s

The object's speed is 10 m/s.

If v instead represented signed velocity in one dimension, both +10 m/s and −10 m/s would be possible.

Physics: Rearranging the pendulum formula

For a simple pendulum swinging through a small angle, its period is approximately:

T = 2π√(L/g)

where:

• T = period (s)

• L = pendulum length (m)

• g = gravitational acceleration (m/s²)

Make L the subject.

Divide by 2π:

T/(2π) = √(L/g)

Square both sides:

T²/(4π²) = L/g

Multiply by g:

L = gT²/(4π²)

Suppose T = 2.0 s and g = 9.8 m/s².

L = 9.8 × 2.0²/(4π²)

L ≈ 0.993 m

The pendulum must be approximately 0.99 m long.

The small-angle pendulum relationship is commonly used in physics experiments to estimate gravitational acceleration.

Geometry: Finding the radius of a sphere

The volume of a sphere is:

V = (4/3)πr³

Make r the subject.

Multiply by 3:

3V = 4πr³

Divide by 4π:

3V/(4π) = r³

Take cube roots:

r = ∛(3V/(4π))

Suppose a spherical container has a volume of 288π cm³.

r = ∛(3 × 288π/(4π))

r = ∛216

r = 6 cm

The radius is 6 cm.

A General Strategy for Rearranging Powers and Roots

When working with a complicated formula, use the following approach:

1. Identify the target variable. Decide which variable must become the subject.

2. Simplify the equation. Combine like terms or clear fractions when useful.

3. Isolate the power or root. Move other terms using inverse operations.

4. Apply the appropriate inverse operation. Use square roots, cube roots, higher roots, or powers.

5. Consider the domain. Decide whether positive, negative, or multiple solutions are possible.

6. Verify the result. Substitute into the original formula and reject any invalid solutions.

Common Mistakes

Forgetting the negative solution

Incorrect:

x² = 36 → x = 6

Correct:

x² = 36 → x = ±6

Both values satisfy the equation, unless the context restricts x.

Taking square roots incorrectly

Incorrect:

√(a² + b²) = a + b

Correct:

√(a² + b²) generally cannot be simplified into separate terms.

For example:

√(9 + 16) = 5, not 3 + 4.

Confusing √x² with x

The correct identity for real x is:

√(x²) = |x|

For example, when x = −5:

√((−5)²) = √25 = 5

This is why absolute values and sign restrictions matter.

Squaring only part of an expression

Incorrect:

(2x + 3)² = 4x² + 9

Correct:

(2x + 3)² = 4x² + 12x + 9

The middle term must be included.

Ignoring extraneous solutions

When an equation contains radicals, solving the squared equation may produce invalid answers. Always substitute possible solutions into the original equation.

Ignoring physical meaning

A negative radius, negative mass, or negative elapsed time is generally not physically meaningful, even if it satisfies an algebraic equation.

Key Terms

• Base: The number or variable being raised to a power.

• Exponent: The number indicating the power to which a base is raised.

• Power: An expression representing exponentiation.

• Square: The result of raising a number to the power of 2.

• Cube: The result of raising a number to the power of 3.

• Square root: A number whose square equals a given non-negative number.

• Cube root: A number whose cube equals a given number.

• Principal square root: The non-negative square root represented by √.

• Radical: An expression involving a root symbol.

• Inverse operations: Operations that reverse one another.

• Subject of a formula: The isolated variable being expressed in terms of other quantities.

• Extraneous solution: A solution produced during manipulation that does not satisfy the original equation.

• Verification: Substituting a solution into the original equation to confirm its validity.

• Domain: The set of allowed input values for an expression or function.

• Fractional exponent: An exponent expressed as a fraction, often representing a combination of powers and roots.

Key Takeaways

• Powers and roots are closely related inverse operations.

• Isolating a squared variable often requires taking square roots of both sides.

• Equations involving even powers can have two real solutions, one solution, or no real solutions.

• Equations of the form xⁿ = a with odd positive integer n have exactly one real solution.

• When rearranging square-root equations, isolate the radical before squaring.

• Squaring can introduce extraneous solutions, so verification is essential.

• Physical contexts can restrict which mathematical solutions are meaningful.

• Fractional exponents provide another way to express roots.

• Correct rearrangement requires preserving equality and respecting the original domain.

Check Your Understanding

Try these questions before looking at the answers.

Recall and Understanding

1. What operation reverses squaring a variable?

2. Explain why x² = 49 has two real solutions.

3. What is the difference between √64 and solving x² = 64?

4. Why can squaring both sides of an equation produce extraneous solutions?

Application

5. Solve: 4x² = 196

6. Solve: (x + 3)² = 81

7. Rearrange A = πr² to make r the subject, assuming r ≥ 0.

8. Solve: 3x³ − 6 = 75

9. Rearrange y = 5√(x + 2) to make x the subject.

10. Solve: x^(3/2) = 64

Analysis and Challenge

11. Solve √(x + 6) = x and check for extraneous solutions.

12. Rearrange KE = ½mv² to make v the subject, assuming v is speed.

13. A sphere has a volume of 500 cm³. Use V = (4/3)πr³ to calculate its radius to two decimal places.

14. Solve x⁴ = 625. Explain why there are two real solutions.

15. A student solves √(x + 9) = x − 3 by squaring both sides and obtains x = 0 and x = 7. Are both values correct? Explain.

Answers to Check Your Understanding

Final thought: Rearranging formulas with powers and roots is not simply about moving symbols around. It requires understanding inverse operations, considering the possible values of variables, and verifying that the resulting solutions satisfy the original relationship.