Algebra 2 Solving Quadratic Equations Test Solutions and Scoring (PDF)

\((x + 2)^{2} = 16\), so \(x + 2 = \pm 4\).

\(x = 2, -6\)

Scoring: 1 point for isolating the square, 1 point for both solutions.

\((3x + 1)(x – 4) = 0\).

\(x = -\frac{1}{3}, 4\)

Scoring: 1 point for the factorization, 1 point for both solutions.

\(x^{2} – 10x = -4\). Add 25: \((x – 5)^{2} = 21\).

\(x = 5 \pm \sqrt{21}\).

\(x = 5 \pm \sqrt{21}\)

Scoring: 1 point for adding 25 to both sides, 1 point for the square, 1 point for the solutions.

Identify \(a = 2, b = -3, c = -4\).

Discriminant: \(b^{2} – 4ac = (-3)^{2} – 4(2)(-4) = 41\).

Substitute: \(x = \frac{-(-3) \pm \sqrt{41}}{2(2)}\).

\(\sqrt{41} = \sqrt{41}\).

x = \(\frac{3 – \sqrt{41}}{4}\) or x = \(\frac{3 + \sqrt{41}}{4}\)

Scoring: 1 point for the substitution, 1 point for the discriminant, 1 point for the simplified solutions.

\(x^{2} – 4 = 0\): \(D = 16 > 0\), two real. \(x^{2} = 0\): \(D = 0\), one real. \(x^{2} + 2 = 0\): \(D = -8 < 0\), none real.

Two, one, none

Scoring: 1 point for the correct counts, 1 point for the discriminants.

Identify \(a = 1, b = 4, c = 8\).

Discriminant: \(b^{2} – 4ac = (4)^{2} – 4(1)(8) = -16\).

Substitute: \(x = \frac{-(4) \pm \sqrt{-16}}{2(1)}\).

\(\sqrt{-16} = 4i\): the solutions are complex.

x = \(-2 + 2i\) or x = \(-2 – 2i\)

Scoring: 1 point for the discriminant, 1 point for the complex solutions.

\(w(w + 5) = 84\), so \(w^{2} + 5w – 84 = 0\), \((w + 12)(w – 7) = 0\).

Only \(w = 7\) is valid.

7 m by 12 m

Scoring: 1 point for the equation, 1 point for the dimensions.

\(-16t^{2} + 32t + 20 = 0\) becomes \(4t^{2} – 8t – 5 = 0\).

\(t = \frac{8 \pm \sqrt{64 + 80}}{8} = \frac{8 \pm 12}{8}\), so \(t = \frac{5}{2}\) or \(t = -\frac{1}{2}\).

At t = 2.5 seconds

Scoring: 1 point for the equation, 1 point for choosing the positive solution.

\(D = 36 – 4k > 0\), so \(4k < 36\).

\(k < 9\)

Scoring: 1 point for the discriminant inequality, 1 point for the answer.