Complex Solutions to Quadratic Equations - College Algebra

1. The quadratic equation $2x^2 - 8x + k = 0$ has complex solutions when the discriminant is negative. If the equation has exactly one complex conjugate pair of solutions, what is the smallest integer value of $k$ that satisfies this condition?
   Answer: 9 For complex solutions, the discriminant $b^2 - 4ac < 0$. Here $a = 2$, $b = -8$, $c = k$. So $(-8)^2 - 4(2)(k) < 0$, which gives $64 - 8k < 0$, so $k > 8$. The smallest integer greater than 8 is 9. Choice A gives $k = 8$, making the discriminant zero (one repeated real solution). Choice C uses $k = 16$ but ignores that we need the smallest integer. Choice D incorrectly calculates the discriminant condition.
2. The equation $2x^2 - 12x + k = 0$ has two complex solutions. If the real parts of both solutions are equal to 3, what is the smallest possible integer value of $k$ such that the solutions are non-real?
   Answer: 19 Using the quadratic formula: $x = \frac{12 \pm \sqrt{144 - 8k}}{4} = \frac{12 \pm \sqrt{144 - 8k}}{4} = 3 \pm \frac{\sqrt{144 - 8k}}{4}$. For the real part to be 3, we need the solutions to be $3 \pm bi$ for some real $b \neq 0$. This means $\frac{\sqrt{144 - 8k}}{4}$ must be purely imaginary, so $144 - 8k < 0$, giving $k > 18$. The smallest integer greater than 18 is 19. Choice A gives $k = 18$, which makes the discriminant zero (one repeated real solution). Choice C gives $k = 17 < 18$. Choice D is larger than necessary.
3. The quadratic $x^2 + px + q = 0$ has solutions $\alpha$ and $\beta$. If $\alpha = 3 - 4i$, and $p$ and $q$ are real, what is the value of $\alpha \cdot \beta$?
   Answer: 25 Since $p$ and $q$ are real, the solutions must be complex conjugates. So $\beta = 3 + 4i$. The product of the solutions is $\alpha \cdot \beta = (3 - 4i)(3 + 4i) = 9 - (4i)^2 = 9 - 16i^2 = 9 + 16 = 25$. Alternatively, by Vieta's formulas, $\alpha \cdot \beta = q$, and we can find $q$ from the factored form: $(x - (3 - 4i))(x - (3 + 4i)) = (x - 3 + 4i)(x - 3 - 4i) = ((x - 3) + 4i)((x - 3) - 4i) = (x - 3)^2 + 16 = x^2 - 6x + 9 + 16 = x^2 - 6x + 25$. So $q = 25$. Choice B uses the sum instead of product. Choice C uses only the real part squared. Choice D uses only the imaginary part squared.
4. If $z = -1 + 2i$ is one solution to the quadratic equation $x^2 + bx + c = 0$ where $b$ and $c$ are real numbers, what is the value of $b + c$?
   Answer: 7 Since the coefficients are real, complex solutions come in conjugate pairs. If $-1 + 2i$ is a solution, then $-1 - 2i$ is also a solution. The quadratic is $(x - (-1 + 2i))(x - (-1 - 2i)) = (x + 1 - 2i)(x + 1 + 2i) = ((x + 1) - 2i)((x + 1) + 2i) = (x + 1)^2 - (2i)^2 = (x + 1)^2 + 4 = x^2 + 2x + 1 + 4 = x^2 + 2x + 5$. So $b = 2$ and $c = 5$, giving $b + c = 7$. Choice A forgets to square the imaginary part correctly. Choice B uses $c$ alone. Choice D uses $b - c$ instead.
5. The polynomial $P(x) = x^2 + 6x + 10$ can be written in the form $P(x) = (x - r_1)(x - r_2)$ where $r_1$ and $r_2$ are complex numbers. What is $|r_1 - r_2|$?
   Answer: 2 Using the quadratic formula: $r_1, r_2 = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2} = -3 \pm i$. So $r_1 = -3 + i$ and $r_2 = -3 - i$. Then $r_1 - r_2 = (-3 + i) - (-3 - i) = 2i$, and $|r_1 - r_2| = |2i| = 2$. Choice B incorrectly uses the distance formula in the complex plane. Choice C doubles the correct answer. Choice D gives the complex number itself rather than its modulus.
6. A quadratic equation of the form $ax^2 + bx + c = 0$ has solutions $x = 2 \pm 3i$. If $a = 1$, what is the value of $c$?
   Answer: 13 If the solutions are $2 + 3i$ and $2 - 3i$, then the quadratic can be written as $a(x - (2 + 3i))(x - (2 - 3i)) = 0$. Expanding: $a((x - 2) - 3i)((x - 2) + 3i) = a((x - 2)^2 - (3i)^2) = a((x - 2)^2 + 9) = a(x^2 - 4x + 4 + 9) = a(x^2 - 4x + 13)$. Since $a = 1$, we have $x^2 - 4x + 13 = 0$, so $c = 13$. Choice A uses only the real part squared. Choice B uses only the imaginary part squared. Choice D incorrectly subtracts instead of adding the squares.

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