Quadratics with Complex Solutions - Math 2

1. If $w = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$ is written in the form $a + bi$, what quadratic equation with integer coefficients has $w$ and $\bar{w}$ as roots?
   Answer: $x^2 - x + 1 = 0$ When you see a complex number in polar form and need to find a quadratic equation with that number and its conjugate as roots, you're working with the relationship between complex conjugate pairs and polynomial coefficients. First, let's convert $w = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$ to rectangular form. Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, we have $w = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. The conjugate is $\bar{w} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. For a quadratic with roots $w$ and $\bar{w}$, we can use the fact that if $r_1$ and $r_2$ are roots, then the quadratic is $x^2 - (r_1 + r_2)x + r_1r_2 = 0$. The sum is: $w + \bar{w} = \frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2} = 1$ The product is: $w\bar{w} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$ Therefore, our quadratic is $x^2 - 1x + 1 = 0$, or $x^2 - x + 1 = 0$. Choice A gives sum = -1 and product = -1. Choice B gives sum = -1 and product = 1. Choice C gives sum = 1 and product = -1. Only choice D matches our calculated sum of 1 and product of 1. Remember: for complex conjugate pairs, their sum is always real (twice the real part) and their product equals the square of the modulus.
2. If $z_1 = 3 + 4i$ and $z_2 = 3 - 4i$ are the roots of a monic quadratic equation $x^2 + px + q = 0$, what quadratic equation has roots $2z_1$ and $2z_2$?
   Answer: $x^2 - 12x + 100 = 0$ First, $2z_1 = 6 + 8i$ and $2z_2 = 6 - 8i$. The sum of the new roots is $(6 + 8i) + (6 - 8i) = 12$, and the product is $(6 + 8i)(6 - 8i) = 36 - (8i)^2 = 36 + 64 = 100$. So the quadratic is $x^2 - 12x + 100 = 0$. Choice B has the wrong sign for the middle term. Choices C and D incorrectly use the original sum and product without scaling.
3. A quadratic equation $ax^2 + bx + c = 0$ has solutions $x = 2 \pm 3i$. If $a = 1$, what is the value of $b + c$?
   Answer: $9$ If the solutions are $x = 2 \pm 3i$, then the quadratic can be written as $(x - (2 + 3i))(x - (2 - 3i)) = 0$. Expanding: $(x - 2 - 3i)(x - 2 + 3i) = (x - 2)^2 - (3i)^2 = (x - 2)^2 - 9i^2 = (x - 2)^2 + 9 = x^2 - 4x + 4 + 9 = x^2 - 4x + 13$. So $b = -4$ and $c = 13$, giving $b + c = 9$.
4. The quadratic equation $ax^2 + bx + c = 0$ has solutions $x = 3 \pm 2i$. If $a = 1$, what is the value of $b^2 - 4ac$?
   Answer: $-16$ Since the solutions are $3 + 2i$ and $3 - 2i$, we can write the quadratic as $a(x - (3 + 2i))(x - (3 - 2i)) = 0$. Expanding: $a((x - 3) - 2i)((x - 3) + 2i) = a((x - 3)^2 - (2i)^2) = a((x - 3)^2 + 4) = a(x^2 - 6x + 9 + 4) = a(x^2 - 6x + 13)$. With $a = 1$, we have $x^2 - 6x + 13 = 0$, so $b = -6$ and $c = 13$. Therefore $b^2 - 4ac = (-6)^2 - 4(1)(13) = 36 - 52 = -16$. Choice B incorrectly takes the absolute value. Choice C squares the imaginary part incorrectly. Choice D combines both errors.
5. For what value of $k$ does the equation $x^2 - 6x + k = 0$ have solutions with imaginary part equal to $\pm 4$?
   Answer: $k = 25$ Using the quadratic formula: $x = \frac{6 \pm \sqrt{36 - 4k}}{2} = 3 \pm \frac{\sqrt{36 - 4k}}{2}$. For the solutions to have imaginary part $\pm 4$, we need $36 - 4k < 0$ (so the discriminant is negative) and $\frac{\sqrt{4k - 36}}{2} = 4$. This gives us $\sqrt{4k - 36} = 8$, so $4k - 36 = 64$, which means $4k = 100$ and $k = 25$. Let's verify: when $k = 25$, we get $x = 3 \pm \frac{\sqrt{-64}}{2} = 3 \pm \frac{8i}{2} = 3 \pm 4i$. Choice A gives imaginary part 0. Choice C gives imaginary part $\pm 2$. Choice D gives imaginary part $\pm 5$.
6. The equation $(x - h)^2 + k^2 = 0$ has solutions $x = 5 \pm 3i$. What is the value of $h + k$?
   Answer: $2$ or $8$ When you encounter an equation involving complex numbers, you need to work backwards from the given solutions to find the parameters. The equation $(x - h)^2 + k^2 = 0$ can be rearranged to $(x - h)^2 = -k^2$, which means $x - h = \pm ki$, so $x = h \pm ki$. Comparing this form with the given solutions $x = 5 \pm 3i$, you can see that $h = 5$ and $ki = 3i$, which means $k = 3$ or $k = -3$ (since $(-3)i = -3i$ and $\pm(-3i) = \mp 3i = \pm 3i$). Therefore, $h + k = 5 + 3 = 8$ or $h + k = 5 + (-3) = 2$. Looking at the answer choices: Choice A gives only 8, missing the possibility that $k = -3$. Choice B incorrectly suggests $h + k$ could be 2 or -8, but -8 is impossible since $h = 5$ (positive) and even if $k = -3$, the sum is still 2. Choice C suggests 8 or -2, but -2 is impossible for the same reason. Choice D correctly identifies both possible values: 2 and 8. The key insight is recognizing that $k$ can be either positive or negative since $k^2$ appears in the original equation, making both values equally valid. Always consider that when you see $k^2$ in an equation, $k$ itself could be positive or negative, leading to multiple valid parameter combinations.
7. If $\alpha$ and $\beta$ are the roots of $x^2 - 4x + 13 = 0$, what is the value of $|\alpha - \beta|$?
   Answer: $6$ using direct calculation methods The roots are $\alpha = \frac{4 + \sqrt{16-52}}{2} = \frac{4 + \sqrt{-36}}{2} = 2 + 3i$ and $\beta = 2 - 3i$. Therefore, $|\alpha - \beta| = |(2+3i) - (2-3i)| = |6i| = 6$. Alternatively, we can use the formula $|\alpha - \beta| = \frac{\sqrt{|\text{discriminant}|}}{|a|} = \frac{\sqrt{36}}{1} = 6$. Choices A, B, and D represent common errors in discriminant calculations or complex number arithmetic.
8. If $z$ is a solution to $2x^2 - 8x + 17 = 0$, what is the value of $|z|^2$?
   Answer: $\frac{17}{2}$ Using the quadratic formula: $z = \frac{8 \pm \sqrt{64 - 136}}{4} = \frac{8 \pm \sqrt{-72}}{4} = \frac{8 \pm 6i\sqrt{2}}{4} = 2 \pm \frac{3i\sqrt{2}}{2}$. For a complex number $a + bi$, we have $|a + bi|^2 = a^2 + b^2$. Here, $a = 2$ and $b = \pm\frac{3\sqrt{2}}{2}$, so $|z|^2 = 2^2 + \left(\frac{3\sqrt{2}}{2}\right)^2 = 4 + \frac{9 \cdot 2}{4} = 4 + \frac{18}{4} = 4 + \frac{9}{2} = \frac{8 + 9}{2} = \frac{17}{2}$. Choice B divides by an extra 2. Choice C uses $5^2$ instead of calculating correctly. Choice D doubles the correct answer.
9. The equation $4x^2 - 8x + 13 = 0$ has solutions $\alpha$ and $\beta$. What is the value of $\frac{1}{\alpha} + \frac{1}{\beta}$?
   Answer: $\frac{8}{13}$ When you encounter a quadratic equation and need to find an expression involving the roots, Vieta's formulas are your most efficient tool. These formulas relate the coefficients of a polynomial to sums and products of its roots without requiring you to solve for the actual roots. For the quadratic $4x^2 - 8x + 13 = 0$ with roots $\alpha$ and $\beta$, Vieta's formulas tell us that $\alpha + \beta = \frac{8}{4} = 2$ and $\alpha\beta = \frac{13}{4}$. To find $\frac{1}{\alpha} + \frac{1}{\beta}$, you need a common denominator: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}$. Substituting the values from Vieta's formulas: $\frac{2}{\frac{13}{4}} = 2 \times \frac{4}{13} = \frac{8}{13}$. Looking at the wrong answers: Choice A gives $\frac{13}{8}$, which results from incorrectly flipping the fraction $\frac{8}{13}$. Choice B shows $-\frac{8}{13}$, which happens if you mistakenly use $\alpha + \beta = -2$ instead of $+2$ (forgetting that for $ax^2 + bx + c = 0$, the sum is $-\frac{b}{a}$, and here $b = -8$). Choice D gives $-\frac{13}{8}$, combining both the sign error and the fraction flip. The key strategy here is recognizing that Vieta's formulas can solve root relationship problems without finding the actual roots. For any quadratic $ax^2 + bx + c = 0$, remember: sum of roots equals $-\frac{b}{a}$ and product equals $\frac{c}{a}$.
10. The quadratic $3x^2 + 12x + 21 = 0$ has complex solutions. If these solutions are written in the form $a \pm bi$ where $a$ and $b$ are real, what is $a^2 + b^2$?
    Answer: $7$ First divide by 3: $x^2 + 4x + 7 = 0$. Using the quadratic formula: $x = \frac{-4 \pm \sqrt{16 - 28}}{2} = \frac{-4 \pm \sqrt{-12}}{2} = \frac{-4 \pm 2i\sqrt{3}}{2} = -2 \pm i\sqrt{3}$. So $a = -2$ and $b = \sqrt{3}$. Therefore $a^2 + b^2 = (-2)^2 + (\sqrt{3})^2 = 4 + 3 = 7$. Choice A uses only $a^2$. Choice C incorrectly calculates $4 + 9 = 13$. Choice D uses $(-4)^2$ instead of $(-2)^2$.
11. If $r$ and $s$ are the complex solutions to $x^2 - 4x + 29 = 0$, what is the value of $r^2 + s^2$?
    Answer: $-42$ When you encounter complex solutions to a quadratic equation, you can often find relationships between the solutions without actually solving for them explicitly. This problem tests your understanding of Vieta's formulas and algebraic manipulation. For the quadratic $x^2 - 4x + 29 = 0$, you can use Vieta's formulas: the sum of roots $r + s = 4$ and the product $rs = 29$. To find $r^2 + s^2$, use the algebraic identity $(r + s)^2 = r^2 + 2rs + s^2$, which rearranges to $r^2 + s^2 = (r + s)^2 - 2rs$. Substituting the values: $r^2 + s^2 = (4)^2 - 2(29) = 16 - 58 = -42$. This confirms answer choice C is correct. Let's examine why the other answers are wrong. Choice A (42) likely comes from incorrectly adding $2rs$ instead of subtracting it: $16 + 26 = 42$. Choice B (-58) might result from forgetting to include the $(r + s)^2$ term entirely and just using $-2rs = -58$. Choice D (58) could come from computing $2rs = 58$ and forgetting about the other terms completely. The key insight here is recognizing that even though the roots are complex, you don't need to find them explicitly. Vieta's formulas combined with algebraic identities let you work with sums and products of roots directly. This approach saves time and reduces calculation errors on exam problems involving complex solutions.
12. If the quadratic $ax^2 + bx + c = 0$ has solutions $x = \frac{1 \pm i\sqrt{3}}{2}$, and $a + b + c = 2$, what is the value of $a$?
    Answer: $a = 2$ from sum of coefficients condition From the solutions, the sum is $\frac{1 + i\sqrt{3}}{2} + \frac{1 - i\sqrt{3}}{2} = 1$ and the product is $\frac{1 + i\sqrt{3}}{2} \cdot \frac{1 - i\sqrt{3}}{2} = \frac{1 - (i\sqrt{3})^2}{4} = \frac{1 + 3}{4} = 1$. So $-\frac{b}{a} = 1$ and $\frac{c}{a} = 1$, giving $b = -a$ and $c = a$. From $a + b + c = 2$: $a + (-a) + a = a = 2$. Choices A, C, and D represent common errors in coefficient relationships or arithmetic.
13. The complex number $z = 2 - 3i$ is a root of the quadratic equation $x^2 + px + q = 0$ where $p$ and $q$ are real. What is another quadratic equation with real coefficients that has $z^2$ as a root?
    Answer: $x^2 + 10x + 169 = 0$ When you see a quadratic with complex roots, remember that complex roots of polynomials with real coefficients always come in conjugate pairs. Since $z = 2 - 3i$ is a root, its conjugate $\overline{z} = 2 + 3i$ must also be a root. To find a quadratic with $z^2$ as a root, first calculate $z^2$: $z^2 = (2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i$ Since we need real coefficients, the conjugate $\overline{z^2} = -5 + 12i$ must also be a root. For a quadratic with roots $r_1$ and $r_2$, the equation is $x^2 - (r_1 + r_2)x + r_1r_2 = 0$. Sum of roots: $(-5 - 12i) + (-5 + 12i) = -10$ Product of roots: $(-5 - 12i)(-5 + 12i) = 25 - (12i)^2 = 25 + 144 = 169$ Therefore: $x^2 - (-10)x + 169 = x^2 + 10x + 169 = 0$ Answer choice A has the wrong sum of roots (5 instead of -10) and wrong product (13 instead of 169). Choice B has the correct product but wrong sum (-10 instead of +10). Choice C has both the wrong sum and wrong product. Only choice D gives us $x^2 + 10x + 169 = 0$. Study tip: When working with powers of complex numbers that need real coefficient polynomials, always find the conjugate of the powered result and use sum/product formulas to build your quadratic.

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