Algebra 2 Complex Conjugates and Complex Solutions Quiz: 10 Questions with Explanations

A \(5 – 3i\)B \(5 + 3i\)C \(3 – 5i\)D \(-5 – 3i\)

Only the imaginary part changes sign. Changing both signs gives 5 – 3i, which is the negative, not the conjugate.

A \(40\)B \(32\)C \(36 – 4i\)D \(36 + 4\)

\(36 – 4i^{2} = 36 + 4 = 40\). A number times its conjugate is a squared plus b squared.

A \(17\)B \(13\)C \(7\)D \(\sqrt{17}\)

Use the Pythagorean theorem: sqrt(25 + 144) = 13. Adding the legs gives 17.

A \(\frac{8}{5} + \frac{4}{5}i\)B \(2 – i\)C \(\frac{8}{5} – \frac{4}{5}i\)D \(\frac{4}{3}\)

Multiply by \(\frac{2 – i}{2 – i}\): numerator 8 – 4i, denominator 4 + 1 = 5.

A \(\frac{9}{2} – \frac{3}{2}i\)B \(\frac{9}{2} + \frac{3}{2}i\)C \(3 + 6i\)D \(9 + 3i\)

\(\frac{(3 + 6i)(1 – i)}{2} = \frac{3 – 3i + 6i + 6}{2} = \frac{9 + 3i}{2}\).

A \(2 \pm 3i\)B \(-2 \pm 3i\)C \(2 \pm 6i\)D \(4 \pm 6i\)

\(x = \frac{4 \pm \sqrt{16 – 52}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i\). Forgetting to halve the 6i gives 2 ± 6i, and using -4 instead of 4 gives the wrong real part.

A \(x = \pm 2i\)B \(x = \pm 2\)C \(x = 4i\)D \(No solutions exist\)

x² = -4, so x = ±2i. The graph shows no real solutions, but complex solutions exist.

A \(-4 + i\)B \(-4 – i\)C \(4 – i\)D \(i – 4\)

Complex solutions of a real polynomial come in conjugate pairs, so the other solution is 4 – i.

A Two real solutionsB One repeated real solutionC No solutions of any kindD Two complex solutions and no real solutions

A negative discriminant means the square root in the quadratic formula is imaginary, so the solutions are complex conjugates.

A \(x^{2} + 6x + 13 = 0\)B \(x^{2} – 6x + 13 = 0\)C \(x^{2} – 6x + 5 = 0\)D \(x^{2} – 3x + 2 = 0\)

Sum of roots 6 and product 9 + 4 = 13 give x² – 6x + 13. A plus sign on the 6x flips the sign of the sum.