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.

Complex roots of real-coefficient quadratics come in conjugate pairs. The conjugate of -4 + 7i is -4 - 7i, formed by negating only the imaginary part. Option A repeats the given root. Option B negates the real part. Option C negates both parts.

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.

Using Vieta's formulas: x² - Sx + P = x² - (-10)x + 34 = x² + 10x + 34 = 0. Option A uses -S = -10 with wrong sign. Options B and D have wrong signs on the product term.

x² + 12x + 36 + 4 = 0, so (x+6)² + 4 = 0, giving (x+6)² = -4. Taking square roots: x+6 = ±2i, yielding x = -6 ± 2i. Option A gives -2, Option C gives -6, Option D gives -8, none of which correctly complete the square for this equation.

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.

D = 6² - 4(3)(13) = 36 - 156 = -120. Option A gives -84 = 36 - 120, using wrong product. Option B gives -100. Option D gives -144. Only -120 correctly applies D = b² - 4ac with a=3, b=6, c=13.

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.

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.

D = 4 - 40 = -36. x = (-2 ± √-36)/2 = (-2 ± 6i)/2 = -1 ± 3i. Option A gives wrong imaginary coefficient 2. Option B gives imaginary coefficient 9. Option C gives wrong real part -2.

True, because when a > 0 the parabola opens upward and if D < 0 there are no real roots, meaning the curve never touches the x-axis and therefore must lie entirely above it for all x, making y always positive.

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.

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.

Sum = (-2+5i) + (-2-5i) = -4. Product = (-2+5i)(-2-5i) = 4 + 25 = 29. Quadratic = x² - (-4)x + 29 = x² + 4x + 29 = 0. Option A uses -4x with wrong sign on sum term. Options B and D have wrong signs on the product.

True, because if D < 0 the quadratic has no real zeros and therefore its graph does not intersect the x-axis at any point, meaning the parabola lies entirely above or below the x-axis depending on the sign of a.