Quadratic Equations with Complex Roots

The solution is ±3i because setting x² + 9 = 0 gives x² = −9 and taking the square root of a negative number introduces i, yielding x = ±√−9 = ±3i, providing two purely imaginary roots.

(4i)² = 16i² = 16(-1) = -16. Option A gives 16, forgetting that i² = -1. Option C gives 8i, halving and leaving imaginary. Option D gives -8i, also incorrect.

A negative discriminant produces a square root involving i, giving solutions of the form a ± bi. These are complex conjugate roots with nonzero imaginary parts. Option A requires positive discriminant. Option B requires zero discriminant. Option D is wrong because complex solutions always exist.

x² = -25, so x = ±√-25 = ±5i. Option A gives real roots but 5² = 25 not -25. Option C gives 25i, using 25 instead of 5 as the coefficient. Option D restates the real root form without the imaginary unit.

(2i)² = 4i² = 4(-1) = -4. Option A forgets to apply i² = -1. Option C gives the result without the sign change from i². Option D halves the result incorrectly.

The statement is true because i³ can be rewritten using powers of i, giving i³ = i² · i, and since i² = −1 by definition of the imaginary unit, substitution yields i³ = (−1) · i = −i, showing that the third power of i simply rotates the unit imaginary number one quarter-turn further around the complex plane, landing at the point −i on the imaginary axis.

True, because any quadratic with real coefficients has roots that are either both real or else a pair of complex conjugates, meaning if 2 + 5i is a root, then 2 − 5i must also be a root so the imaginary parts cancel in coefficient expressions.

True, because the discriminant b² − 4ac completely determines the nature of the roots: positive gives two real roots, zero gives one repeated real root, and negative gives two complex conjugate roots.

The roots are 2 ± 3i because the discriminant is −36 and applying the quadratic formula gives x = (4 ± 6i)/2 = 2 ± 3i, forming a conjugate pair.

The correct statement is that a negative discriminant produces two complex conjugate roots because when D < 0 the square root term becomes imaginary, giving solutions a ± bi which always occur in conjugate pairs for quadratics with real coefficients.

The correct value is −1 because i is defined as √−1, so squaring it gives i² = (√−1)² = −1, which is the fundamental identity used in all complex-number algebra.

(2i)² equals −4 because (2i)² = 4i² and i² = −1, so the result is 4(−1) = −4, converting a squared imaginary term into a negative real number.

False, because complex numbers require a two-dimensional representation in the Argand plane where the real part lies on the horizontal axis and the imaginary part on the vertical axis, making them impossible to plot on a one-dimensional number line.

The product (3 + 2i)(3 − 2i) equals 13 because multiplying a complex number by its conjugate yields a² + b², so here 3² + 2² = 9 + 4 = 13, and the imaginary parts cancel to produce a purely real result.

False, because a discriminant of zero means the quadratic has exactly one real repeated root, not a pair of complex roots, since the square root of zero introduces no imaginary term.

True, because the sum of complex conjugate roots (a + bi) and (a − bi) is 2a, which is always real as the imaginary parts cancel perfectly.

The roots are −2 ± 2i because applying the quadratic formula gives x = (−4 ± √−16)/2 = (−4 ± 4i)/2 = −2 ± 2i, showing a pair of complex conjugate solutions.

The discriminant is −16 because D = b² − 4ac = 4² − 4·1·8 = 16 − 32 = −16, and since the discriminant is negative the quadratic has complex (not real) roots.

By Vieta's formulas, sum of roots = -b/a = -6/1 = -6. Option A gives +6, the wrong sign. Option C gives the product of the roots not the sum. Option D negates the product.

By Vieta's formulas, product of roots = c/a = 13/1 = 13. This is real because multiplying conjugate roots (a+bi)(a-bi) = a² + b² always gives a positive real value. Option A gives the negative of b, option B negates it, option D negates 13.