This is a complex roots question testing the quadratic formula with a negative discriminant. Choice A (2 ± 3i) is correct — apply the quadratic formula: x = (4 ± √(16 − 52)) / 2 = (4 ± √(−36)) / 2 = (4 ± 6i) / 2 = 2 ± 3i. The discriminant is 16 − 52 = −36, which is negative, confirming complex (non-real) roots. Choice B (−2 ± 3i) correctly computes √(−36) = 6i and divides by 2, but uses −b instead of b/2a for the real part: applying −(−4)/2 = −2 rather than 4/2 = 2. Choice C (4 ± 6i) correctly computes the numerator 4 ± 6i but forgets to divide by 2 (the denominator of the quadratic formula). Choice D states "no complex roots" — this contradicts the fact that a negative discriminant guarantees two complex conjugate roots. Pro tip: A negative discriminant (b² − 4ac < 0) means the quadratic has complex roots, not "no roots." Write √(−36) = √36 · √(−1) = 6i, then divide the entire numerator by 2a.

The correct answer is A (1 + 2i). Multiply by the conjugate of the denominator: (3 + i)/(1 − i) × (1 + i)/(1 + i). Numerator: (3 + i)(1 + i) = 3 + 3i + i + i² = 3 + 4i − 1 = 2 + 4i. Denominator: (1 − i)(1 + i) = 1 − i² = 1 + 1 = 2. Result: (2 + 4i)/2 = 1 + 2i. B (2 + i) likely comes from incomplete multiplication or forgetting to divide. C (1 − 2i) comes from a sign error in the numerator expansion. D (2 − i) comes from multiplying by the wrong conjugate (1 − i) instead of (1 + i). The key technique: always multiply by the conjugate to eliminate i from the denominator, remembering that i² = −1.

This problem asks for the complex conjugate of $-6 + 7i$, which is found by changing the sign of the imaginary part only. The real part is $-6$, and the imaginary part $7i$ becomes $-7i$. Thus, the conjugate is $-6 - 7i$. Choice D might result from incorrectly flipping the sign of the real part instead of the imaginary part.

Subtracting complex numbers requires distributing the negative sign and combining like terms. $(8 + 5i) - (3 - 9i) = 8 + 5i - 3 + 9i$. Combining real parts: $8 - 3 = 5$, and combining imaginary parts: $5i + 9i = 14i$. The result in standard form is $5 + 14i$.

Adding complex numbers involves combining real and imaginary parts separately. For $(2 + 3i) + (7 - 10i)$, we add the real parts: $2 + 7 = 9$, and add the imaginary parts: $3i + (-10i) = -7i$. The result in standard form is $9 - 7i$. This represents combining two complex impedances in an AC circuit.

Using FOIL to multiply: $(x + 2i)(3 - i) = x \cdot 3 + x \cdot(-i) + (2i) \cdot 3 + (2i) \cdot(-i) = 3x - xi + 6i - 2i^2$. Since $i^2 = -1$, this becomes $3x - xi + 6i - 2(-1) = 3x - xi + 6i + 2$. The real part consists of terms without $i$: $3x + 2$.

The magnitude formula is $|a + bi| = \sqrt{a^2 + b^2}$. For $(3 - 4i)$, we have $a = 3$ and $b = -4$, so the magnitude is $\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. This represents the distance from the origin to the point $(3, -4)$ in the complex plane, forming a 3-4-5 right triangle.

This problem requires multiplying two complex numbers using the FOIL method. (1 + 6i)(4 - i) = 1(4) + 1(-i) + 6i(4) + 6i(-i) = 4 - i + 24i - 6i². Since i² equals negative one, this becomes 4 + 23i - 6(-1) = 4 + 23i + 6 = 10 + 23i. Choice C likely forgot that i² equals negative one.

This is multiplication of complex numbers using FOIL. $$ (3 + 5i)(2 - 3i) = 3 \cdot 2 + 3 \cdot(-3i) + 5i \cdot 2 + 5i \cdot(-3i) = 6 - 9i + 10i - 15i^2 $$. Since $i^2 = -1$, this becomes $6 + i - 15(-1) = 6 + i + 15 = 21 + i$. The real part is $21$.

This is multiplication of complex numbers using FOIL. $(2 + i)(3 - i) = 2 \cdot 3 + 2 \cdot(-i) + i \cdot 3 + i \cdot(-i) = 6 - 2i + 3i - i^2$. Since $i^2 = -1$, this becomes $6 + i - (-1) = 6 + i + 1 = 7 + i$.