Algebra 2 Radical and Rational Equations Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Radical and Rational Equations – Maths PDF

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This radical rational equations answers for Algebra 2 covers radical and rational equations (Common Core A-REI.A.2), with a free PDF to print.

Every worksheet problem on radical and rational equations is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where keeping an extraneous solution can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: A-REI.A.2

Radical rational equations answers: exercises 1 to 10

Exercise 1 – Basic radical equation

Square both sides: \(x + 5 = 16\), so \(x = 11\).

Check: \(\sqrt{16} = 4\).

\(x = 11\)

Exercise 2 – Radical with a coefficient

\(2x – 1 = 25\), so \(x = 13\). Check: \(\sqrt{25} = 5\).

\(x = 13\)

Exercise 3 – Extraneous solution

One intersection at x = 3.

\(x + 6 = x^{2}\), so \(x^{2} – x – 6 = 0\) and \((x – 3)(x + 2) = 0\).

Check \(x = 3\): \(\sqrt{9} = 3\), true. Check \(x = -2\): \(\sqrt{4} = 2 \ne -2\), extraneous.

\(x = 3\)

Exercise 4 – Radical equal to a binomial

Square: \(x^{2} – 2x + 1 = x + 11\), so \(x^{2} – 3x – 10 = 0\) and \((x – 5)(x + 2) = 0\).

Check \(x = 5\): \(4 = \sqrt{16}\), true. Check \(x = -2\): \(-3 = \sqrt{9} = 3\), false.

\(x = 5\)

Exercise 5 – Both solutions valid

Both candidates are valid.

Isolate: \(\sqrt{3x + 4} = x + 2\). Square: \(3x + 4 = x^{2} + 4x + 4\), so \(x^{2} + x = 0\) and \(x = 0\) or \(-1\).

Check \(x = 0\): \(\sqrt{4} – 2 = 0\). Check \(x = -1\): \(\sqrt{1} – 2 = -1\). Both are valid.

\(x = 0\) and \(x = -1\)

Exercise 6 – Cube root equation

Cube both sides: \(x – 3 = 8\).

\(x = 11\)

Exercise 7 – Radicals on both sides

Square: \(x + 7 = 2x – 1\), so \(x = 8\). Check: \(\sqrt{15} = \sqrt{15}\).

\(x = 8\)

Exercise 8 – Two radicals

Isolate one radical: \(\sqrt{x + 3} = 3 – \sqrt{x}\). Square: \(x + 3 = 9 – 6\sqrt{x} + x\).

So \(6\sqrt{x} = 6\), \(\sqrt{x} = 1\), \(x = 1\). Check: \(2 + 1 = 3\).

\(x = 1\)

Exercise 9 – Proportion

Cross multiply: \(3x = 24\).

\(x = 8\)

Exercise 10 – Rational equation

Cross multiply: \(3x = x + 2\), so \(x = 1\). Excluded values are \(-2\) and \(0\).

\(x = 1\)

Exercises 11 and up

Exercise 11 – Clear the fractions

Multiply by the LCD \(3x\): \(6 + x = 3x\), so \(x = 3\).

Check: \(\frac{2}{3} + \frac{1}{3} = 1\).

\(x = 3\)

Exercise 12 – An extraneous solution

The candidate x = 2 is excluded.

Multiply by \(x – 2\): \(x = 2 + 3x – 6\), so \(-2x = -4\) and \(x = 2\).

But \(x = 2\) makes the denominator 0, so it is extraneous.

No solution

Exercise 13 – Common denominator

Multiply by \(x^{2} – 1\): \((x + 1) + (x – 1) = 4\), so \(2x = 4\) and \(x = 2\).

\(x = 2\)

Exercise 14 – Both candidates excluded

Multiply by \((x + 3)(x – 3)\): \(x(x – 3) + 3(x + 3) = 18\), so \(x^{2} + 9 = 18\) and \(x = \pm 3\).

Both values make a denominator 0.

No solution

Exercise 15 – Rational leads to a quadratic

Multiply by \(x\): \(6 = x^{2} – x\), so \(x^{2} – x – 6 = 0\) and \((x – 3)(x + 2) = 0\).

Check \(x = 3\): \(2 = 2\). Check \(x = -2\): \(-3 = -3\).

\(x = 3\) and \(x = -2\)

Exercise 16 – Filling a pool

Rates: \(\frac{1}{t} + \frac{1}{t + 3} = \frac{1}{2}\). Multiply by \(2t(t + 3)\): \(2(t + 3) + 2t = t(t + 3)\).

\(4t + 6 = t^{2} + 3t\), so \(t^{2} – t – 6 = 0\) and \((t – 3)(t + 2) = 0\). Time cannot be negative.

Hose A: 3 hours, hose B: \(3 + 3 = 6\) hours. Check: \(\frac{1}{3} + \frac{1}{6} = \frac{1}{2}\).

Rates 1/3, 1/6, 1/2.

Hose A: 3 hours; hose B: 6 hours

Exercise 17 – Boat and current

Check: 24/8 + 24/12 = 3 + 2 = 5.

Multiply by \((v – 2)(v + 2)\): \(24(v + 2) + 24(v – 2) = 5(v^{2} – 4)\), so \(48v = 5v^{2} – 20\).

\(5v^{2} – 48v – 20 = 0\) factors as \((5v + 2)(v – 10) = 0\). Speed is positive, so \(v = 10\).

Check: \(\frac{24}{8} + \frac{24}{12} = 3 + 2 = 5\).

10 mph

Exercise 18 – Free fall

\(\frac{\sqrt{h}}{4} = 3\), so \(\sqrt{h} = 12\). Square: \(h = 144\).

144 feet

Exercise 19 – Skid marks

\(\sqrt{24d} = 36\). Square: \(24d = 1296\), so \(d = 54\).

54 feet

Exercise 20 – Find the error

Squaring both sides can create extraneous solutions. For \(x = -2\) the left side is \(\sqrt{4} = 2\), which is not \(-2\).

x = -2 is extraneous; the only solution is x = 3

Key facts behind this radical rational equations answers

Key facts

Isolate the radical before squaring.
Squaring can create extraneous solutions; always check.
Rational equations: multiply by the LCD to clear fractions.
Reject answers that make a denominator 0.
A principal square root is never negative.

How to use this radical rational equations answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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