Algebra 2 Complex Conjugates and Complex Solutions Test Solutions and Scoring (PDF)

\(z = 3 – 4i\), conjugate \(3 + 4i\), \(|z| = \sqrt{9 + 16} = 5\).

\(3 – 4i\); \(3 + 4i\); \(5\)

Scoring: 1 point for z and the conjugate, 1 point for the modulus.

1. Multiply the numerator and denominator by the conjugate of the denominator, \(1 + i\):

   \(\frac{(7 + i)(1 + i)}{(1 – i)(1 + i)}\)

   Numerator: \(6 + 8i\). Denominator: \(1 + 1 = 2\).

   Divide each part by \(2\): \(3 + 4i\).

   \(3 + 4i\)

2. Multiply the numerator and denominator by the conjugate of the denominator, \(-i\):

   \(\frac{(2 – 3i)(-i)}{(i)(-i)}\)

   Numerator: \(-3 – 2i\). Denominator: \(0 + 1 = 1\).

   Divide each part by \(1\): \(-3 – 2i\).

   \(-3 – 2i\)

Scoring: Part (a): 1 point for multiplying by the conjugate, 1 point for the answer. Part (b): 1 point.

Identify \(a = 1, b = -2, c = 5\).

Discriminant: \(b^{2} – 4ac = (-2)^{2} – 4(1)(5) = -16\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{2 \pm \sqrt{-16}}{2}\) with \(\sqrt{-16} = 4i\).

\(x = \frac{2 \pm 4i}{2} = 1 \pm 2i\).

Second equation: \(x^{2} = -16\), so \(x = \pm 4i\).

\(1 \pm 2i\); \(\pm 4i\)

Scoring: First equation: 1 point for the discriminant, 1 point for both solutions. Second equation: 1 point.

The vertex (-3, 2) is above the x-axis and the parabola opens up, so there are 0 real solutions.

\(b^{2} – 4ac = 36 – 44 = -8 < 0\), which confirms two complex solutions.

0 real solutions; discriminant -8

Scoring: 1 point for reading the graph, 1 point for the discriminant.

Sum \(= 2\). Product \(= 1 + 16 = 17\).

\(x^{2} – 2x + 17 = 0\)

Scoring: 1 point for the sum and product, 1 point for the equation.

Multiply the numerator and denominator by the conjugate of the denominator, \(2 – i\):

\(\frac{(12 + 16i)(2 – i)}{(2 + i)(2 – i)}\)

Numerator: \(40 + 20i\). Denominator: \(4 + 1 = 5\).

Divide each part by \(5\): \(8 + 4i\).

\(8 + 4i\) ohms

Scoring: 1 point for the conjugate step, 1 point for the answer.

\((1 + 2i)(1 – 2i) = 1 + 4 = 5\). \(|z|^{2} = (\sqrt{5})^{2} = 5\).

Both equal 5

Scoring: 1 point for the product, 1 point for the comparison.

\(-16t^{2} + 48t – 40 = 0\) simplifies to \(2t^{2} – 6t + 5 = 0\). Discriminant: \(36 – 40 = -4\).

No: the discriminant is negative (max height 36 ft)

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

Identify \(a = 3, b = -6, c = 7\).

Discriminant: \(b^{2} – 4ac = (-6)^{2} – 4(3)(7) = -48\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{6 \pm \sqrt{-48}}{6}\) with \(\sqrt{-48} = 4\sqrt{3}i\).

\(x = \frac{6 \pm 4i\sqrt{3}}{6}\) reduces to \(1 \pm \frac{2\sqrt{3}}{3}i\).

\(x = 1 \pm \frac{2\sqrt{3}}{3}i\)

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