Algebra 2 Radical and Rational Equations Quiz: 10 Questions with Explanations

A \(x = 12\)B \(x = 36\)C \(x = \pm 36\)D \(x = 6\)

Square both sides: x = 36. The principal root is nonnegative, so there is only one solution.

A A solution that is a fractionB A value that solves the transformed equation but not the originalC A solution that is negativeD A solution found by factoring

Extraneous solutions are introduced by steps such as squaring or multiplying by an expression that could be zero.

A 0B 2C 3D 1

The graphs cross only once, at (3, 3). The value x = -2 from squaring is extraneous.

A \(x = 10\)B \(x = 1.6\)C \(x = 0.1\)D \(x = 8\)

Cross multiply: 2x = 20, so x = 10.

A \(x = 0\)B \(x = 3\)C \(x = 5\)D \(x = -5\)

x = 5 makes the denominators zero.

A \(x = 4\)B \(x = 6\)C \(x = 10\)D \(x = 26\)

Isolate: √(x – 1) = 3, so x – 1 = 9 and x = 10.

A Check for excluded valuesB Square both sidesC Add the denominatorsD Take the reciprocal

The last step is checking that no solution makes a denominator equal to zero.

A \(x = -8\)B \(x = 8\)C \(No solution\)D \(x = -6\)

Cube both sides: x = (-2)³ = -8. Cube roots of negative numbers are real.

A \(x = \frac{1}{2}\)B \(x = 4\)C \(x = 3\)D \(x = 2\)

1/x + 1/(2x) = 3/(2x) = 3/4 gives x = 2.

A Square both sides immediatelyB Isolate the radicalC Multiply by the LCDD Add 3 to both sides of only one term

Isolate the radical first: √(x + 2) = x – 2. Squaring before isolating makes the algebra harder.