Quadratic Equations with Complex Roots

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.

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.

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.

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 correct statement is that a negative discriminant produces two complex conjugate roots because when D

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.

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.

(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.

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.

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.

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.

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.

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 imaginary unit i is defined as the square root of −1 because this definition enables expression of square roots of negative numbers and forms the foundation for all complex arithmetic.

The solutions are ±5i because x² + 25 = 0 implies x² = −25 and taking square roots gives x = ±√−25 = ±5i, producing two imaginary roots.

The sum of the roots of x² + 6x + 13 = 0 is −6 because for any quadratic ax² + bx + c the sum of roots is −b/a, and here −6/1 = −6.

The product of the roots is 13 because in a quadratic ax² + bx + c the product of the roots equals c/a, so here c = 13 and a = 1, giving a product of 13 which is real as it comes from multiplying conjugate roots.

When b² − 4ac

(4i)² equals −16 because 4i multiplied by itself yields 16i² and i² = −1, giving the final value 16(−1) = −16.