Quadratic with Complex (Advanced)

Dividing the equation by 2 gives x² + 2x + 4 = 0, whose discriminant is D = 2² − 4·1·4 = 4 − 16 = −12, and applying the quadratic formula yields x = (−2 ± √−12)/2 = (−2 ± i√12)/2 = −1 ± i√3, so the roots are a pair of complex conjugates formed by completing the square or directly evaluating the negative discriminant.

The discriminant of the quadratic is D = 4² − 4·4·5 = 16 − 80 = −64, and substituting this into the quadratic formula gives x = (−4 ± √−64)/8 = (−4 ± 8i)/8 = −½ ± i, so the equation has two complex roots obtained by taking the square root of −64 and simplifying the fractional form.

Since the conjugate of the given root 3 − 2i is 3 + 2i, their sum is 6 and their product is (3 − 2i)(3 + 2i) = 9 + 4 = 13, and using Vieta’s formulas the quadratic with these roots is x² − (sum)x + product = x² − 6x + 13 = 0, which is the unique monic quadratic whose real coefficients generate the required conjugate pair.

Completing the square transforms the quadratic x² + 10x + 29 into (x + 5)² + 4 by adding and subtracting 25 inside the expression, and setting it equal to zero gives (x + 5)² = −4, which shows the squared term must equal a negative number, so taking square roots introduces the imaginary unit and yields x + 5 = ±2i, leading to the final complex conjugate roots x = −5 ± 2i.

The correct statement is C because computing the discriminant gives D = (−6)² − 4·1·13 = 36 − 52 = −16, and since the discriminant is negative the quadratic has no real solutions and must instead have two complex conjugate roots.

The quadratic with roots 2 + i and 2 − i is x² − 4x + 5 = 0 because the sum of the roots is (2 + i) + (2 − i) = 4 and their product is (2 + i)(2 − i) = 4 + 1 = 5, and substituting these into the standard form x² − (sum)x + (product) produces x² − 4x + 5 = 0.

The product (5 − 2i)(5 + 2i) equals 29 because multiplying a complex number by its conjugate yields a² + b², so here 5² + 2² = 25 + 4 = 29, and the imaginary parts cancel completely leaving a real result.

The pair (sum of roots, product of roots) is (−8, 25) because for any quadratic ax² + bx + c the sum of the roots is −b/a = −8 and the product is c/a = 25, and even when the roots are complex these values remain real because they depend only on the real coefficients.

True, because if D

True, because for any quadratic with real coefficients a root of the form a + bi must be paired with its conjugate a − bi so that the imaginary parts cancel in the expanded polynomial, ensuring all coefficients remain real.

True, because if (x − h)² = −k with k > 0 then taking square roots gives x − h = ±√(−k) = ±i√k, making the solutions complex since the square root of a negative number necessarily introduces the imaginary unit i.

True, because multiplying a complex number a + bi by its conjugate a − bi yields a² + b², which is always a non-negative real number since both squared terms are ≥ 0 and their sum is zero only when a = b = 0, meaning the original number was 0.

False, because even if a quadratic has complex roots its vertex value y = c − b²/(4a) depends only on real coefficients a, b, and c, so the vertex’s y-coordinate is always a real number regardless of whether the roots are real or complex.

True, because when a > 0 the parabola opens upward and if D

The solution to x² + 2x + 10 = 0 is x = −1 ± 3i because the discriminant is D = 4 − 40 = −36 and applying the quadratic formula gives x = (−2 ± √−36)/2 = (−2 ± 6i)/2 = −1 ± 3i, yielding a complex conjugate pair.

The discriminant of 3x² + 6x + 13 = 0 is −120 because D = 6² − 4·3·13 = 36 − 156 = −120, a negative value indicating two complex roots.

Completing the square for x² + 12x + 40 gives (x + 6)² = −4 because rewriting x² + 12x + 36 + 4 = 0 leads to (x + 6)² + 4 = 0 and therefore (x + 6)² = −4, producing roots x = −6 ± 2i.

A quadratic with sum of roots S = −10 and product P = 34 is x² + 10x + 34 = 0 because the standard form using Vieta’s formulas is x² − Sx + P, and substituting S = −10 and P = 34 yields x² − (−10)x + 34 = x² + 10x + 34 = 0.

If one root is −4 + 7i then the other is −4 − 7i because any quadratic with real coefficients must have complex conjugate roots so that the imaginary parts cancel out in the polynomial coefficients.

The quadratic with roots −2 ± 5i is x² + 4x + 29 = 0 because the sum of the roots is −4 and the product is (−2 + 5i)(−2 − 5i) = 4 + 25 = 29, and substituting into x² − (sum)x + product yields x² − (−4)x + 29 = x² + 4x + 29 = 0.