Algebra 2 Solving Quadratic Equations Quiz: 10 Questions with Explanations

A \(x = 9\)B \(x = \pm 9\)C \(x = \pm 40.5\)D \(x = 81\)

Square roots give both a positive and a negative solution.

A \(x = -2 and x = 7\)B \(x = 2 and x = 7\)C \(x = 2 and x = -7\)D \(x = 14\)

Each factor set to 0: x – 2 = 0 gives 2 and x + 7 = 0 gives -7.

A \(12\)B \(-12\)C \(44\)D \(-9\)

b² – 4ac = 16 – 28 = -12, so there are two complex solutions.

A The parabola that stays above the x-axisB The parabola that touches the x-axis onceC The parabola that crosses the x-axis twiceD None of them

A negative discriminant means no real solutions, so the parabola never meets the x-axis.

A \(x = -2 \pm \sqrt{5}\)B \(x = 2 \pm \sqrt{5}\)C \(x = -2 \pm \sqrt{3}\)D \(x = -4 \pm \sqrt{5}\)

x² + 4x + 4 = 5, so (x + 2)² = 5 and x = -2 ± √5.

A \(x = -4 and x = 1\)B \(x = 4 and x = -1\)C \(x = 3 and x = 4\)D \(x = 4 and x = 1\)

D = 9 + 16 = 25, x = (3 ± 5)/2 = 4 or -1.

A \(w^{2} + 2w + 24 = 0\)B \(2w + 2 = 24\)C \(w^{2} – 2w – 24 = 0\)D \(w^{2} + 2w – 24 = 0\)

Area = w(w + 2) = 24, so w² + 2w – 24 = 0, which gives w = 4.

A OneB TwoC ThreeD None

D = 9 – 40 = -31 < 0, so there are no real solutions.

A \(x^{2} – 6x + 5 = 0\)B \(x^{2} + 6x + 5 = 0\)C \(x^{2} – 5 = 0\)D \(x^{2} – 6x – 5 = 0\)

Solutions 1 and 5 have sum 6 and product 5, so x² – 6x + 5 = 0.

A \(x = \pm 5\)B \(x = 25i\)C \(x = \pm 5i\)D \(No solution\)

x² = -25, so x = ±5i. There are no real solutions but there are two complex ones.