SAT — mathHard5 min read

Solving Radical and Rational Equations

Solve equations with square roots and algebraic fractions for the Digital SAT. Learn to eliminate radicals and check for extraneous solutions.

Tutor AI Team
Tutor AI Team

Radical equations contain variables under a radical sign (\sqrt{}), and rational equations contain variables in the denominator. Both require special solving techniques and both can produce extraneous solutions — values that satisfy the transformed equation but not the original. The Digital SAT tests both types.

Core Concepts

Solving Radical Equations

  1. Isolate the radical on one side.
  2. Square both sides to eliminate the square root.
  3. Solve the resulting equation.
  4. Check for extraneous solutions by substituting back.

Example: x+3=5\sqrt{x + 3} = 5

Square: x+3=25x + 3 = 25

x=22x = 22

Check: 22+3=25=5\sqrt{22 + 3} = \sqrt{25} = 5 ✓

Why Extraneous Solutions Occur

Squaring both sides can introduce false solutions because squaring is not a reversible operation ((−5)2=25(-5)^2 = 25 and 52=255^2 = 25).

Example: x=−3\sqrt{x} = -3

Square: x=9x = 9

Check: 9=3≠−3\sqrt{9} = 3 \neq -3 ✗. Extraneous! No solution.

Radical Equations with Two Radicals

Isolate one radical, square, then isolate the remaining radical and square again.

Solving Rational Equations

  1. Find the LCD of all fractions.
  2. Multiply every term by the LCD to clear denominators.
  3. Solve the resulting equation.
  4. Check for extraneous solutions — solutions that make any original denominator zero.

Example: 5x+13=2x\frac{5}{x} + \frac{1}{3} = \frac{2}{x}

LCD = 3x3x. Multiply through:

15+x=615 + x = 6

x=−9x = -9

Check: 5−9+13=−59+39=−29\frac{5}{-9} + \frac{1}{3} = -\frac{5}{9} + \frac{3}{9} = -\frac{2}{9} and 2−9=−29\frac{2}{-9} = -\frac{2}{9} ✓

Cross-Multiplying

When the equation has one fraction on each side:

ab=cd  ⟹  ad=bc\frac{a}{b} = \frac{c}{d} \implies ad = bc

Example: x+13=x−25\frac{x+1}{3} = \frac{x-2}{5}

5(x+1)=3(x−2)5(x+1) = 3(x-2) → 5x+5=3x−65x + 5 = 3x - 6 → 2x=−112x = -11 → x=−112x = -\frac{11}{2}

Strategy Tips

Tip 1: Always Check Your Answers

Both radical and rational equations can produce extraneous solutions. Never skip the check step.

Tip 2: Isolate the Radical Before Squaring

If you square without isolating, you get a more complex equation. x+1+2=5\sqrt{x + 1} + 2 = 5 → first get x+1=3\sqrt{x+1} = 3, then square.

Tip 3: Watch for No Solution

A square root cannot equal a negative number. If you get expression=negative\sqrt{\text{expression}} = \text{negative}, there's no solution.

Tip 4: For Rational Equations, State the Restrictions

Before solving, note which values make denominators zero. If your solution is one of these, it's extraneous.

Tip 5: Use the Desmos Calculator

Graph both sides as separate functions. The x-coordinates of intersections are the solutions.

Worked Example: Example 1

Problem

Solve 3x+1=4\sqrt{3x + 1} = 4.

Square: 3x+1=163x + 1 = 16 → 3x=153x = 15 → x=5x = 5

Check: 16=4\sqrt{16} = 4 ✓

Solution

Worked Example: Example 2

Problem

Solve 2x+7=x+2\sqrt{2x + 7} = x + 2.

Square: 2x+7=(x+2)2=x2+4x+42x + 7 = (x + 2)^2 = x^2 + 4x + 4

0=x2+2x−3=(x+3)(x−1)0 = x^2 + 2x - 3 = (x + 3)(x - 1)

x=−3x = -3 or x=1x = 1

Check x=−3x = -3: 1=1\sqrt{1} = 1 but −3+2=−1-3 + 2 = -1. 1≠−11 \neq -1 ✗ Extraneous.

Check x=1x = 1: 9=3\sqrt{9} = 3 and 1+2=31 + 2 = 3 ✓

Solution: x=1x = 1

Solution

Worked Example: Example 3

Problem

Solve 2x−1=6x+3\frac{2}{x - 1} = \frac{6}{x + 3}.

Cross-multiply: 2(x+3)=6(x−1)2(x + 3) = 6(x - 1)

2x+6=6x−62x + 6 = 6x - 6 → 12=4x12 = 4x → x=3x = 3

Check: 22=1\frac{2}{2} = 1 and 66=1\frac{6}{6} = 1 ✓

Solution

Worked Example: SAT-Style

Problem

Solve xx−2+1x+2=8x2−4\frac{x}{x-2} + \frac{1}{x+2} = \frac{8}{x^2-4}.

Note: x2−4=(x−2)(x+2)x^2 - 4 = (x-2)(x+2). LCD = (x−2)(x+2)(x-2)(x+2).

Multiply through: x(x+2)+1(x−2)=8x(x+2) + 1(x-2) = 8

x2+2x+x−2=8x^2 + 2x + x - 2 = 8

x2+3x−10=0x^2 + 3x - 10 = 0

(x+5)(x−2)=0(x + 5)(x - 2) = 0 → x=−5x = -5 or x=2x = 2

x=2x = 2 makes the original denominator 0 → extraneous.

Solution: x=−5x = -5

Solution

Worked Example: Example 5

Problem

Solve x+5+1=x\sqrt{x + 5} + 1 = x.

Isolate: x+5=x−1\sqrt{x + 5} = x - 1

Square: x+5=x2−2x+1x + 5 = x^2 - 2x + 1

x2−3x−4=0x^2 - 3x - 4 = 0 → (x−4)(x+1)=0(x - 4)(x + 1) = 0

x=4x = 4: 9+1=4\sqrt{9} + 1 = 4 ✓

x=−1x = -1: 4+1=3≠−1\sqrt{4} + 1 = 3 \neq -1 ✗

Solution: x=4x = 4

Solution

Practice Problems

  1. Problem 1

    Solve 5x−1=7\sqrt{5x - 1} = 7.

    Problem 2

    Solve x+4=x−2\sqrt{x + 4} = x - 2.

    Problem 3

    Solve 3x+14=1x\frac{3}{x} + \frac{1}{4} = \frac{1}{x}.

    Problem 4

    Solve x+1x−3=4x−3\frac{x + 1}{x - 3} = \frac{4}{x - 3}.

    Problem 5

    Solve 2x+3=−1\sqrt{2x + 3} = -1.

    Problem 6

    Solve 1x−1+1x+1=4x2−1\frac{1}{x-1} + \frac{1}{x+1} = \frac{4}{x^2-1}.

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Common Mistakes

  • Not checking for extraneous solutions. This is critical — always substitute back.
  • Squaring without isolating the radical. (x+2)2≠x+4(\sqrt{x} + 2)^2 \neq x + 4. It equals x+4x+4x + 4\sqrt{x} + 4.
  • Forgetting that x\sqrt{x} can't equal a negative. If isolating gives expr=negative\sqrt{\text{expr}} = \text{negative}, stop — no solution.
  • Cancelling denominators without checking restrictions. Note excluded values before solving.
  • Arithmetic errors when squaring binomials. (x+2)2=x2+4x+4(x + 2)^2 = x^2 + 4x + 4, not x2+4x^2 + 4.

Frequently Asked Questions

Why do extraneous solutions appear?

Because squaring both sides (or multiplying by a variable expression) can introduce solutions that don't satisfy the original equation.

Is there a way to avoid extraneous solutions?

Not entirely — the method inherently may produce them. Just always check.

How common are these on the SAT?

1–2 questions per test. Radical equations are slightly more common than rational equations.

Can I use Desmos to solve these?

Yes — graph both sides and find intersections.

What if the equation has two radicals?

Isolate one, square, simplify, then isolate the remaining radical and square again.

Key Takeaways

  • ✓

    Radical equations: isolate → square → solve → CHECK.

  • ✓

    Rational equations: find LCD → multiply through → solve → CHECK.

  • ✓

    Extraneous solutions can and do occur — never skip the check step.

  • ✓

    expression≥0\sqrt{\text{expression}} \geq 0 always — a radical can't equal a negative.

  • ✓

    Cross-multiply when you have one fraction on each side.

  • ✓

    State restrictions for rational equations before solving.

Tags:
#sat#math#advanced-math#radical-equations#rational-equations

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