Complex Roots of Quadratics - Algebra 3

1. The equation $2x^2 + px + q=0$, where $p$ and $q$ are real, has $3-i$ as a root. What is $p+q$?
   Answer: 8 Because the coefficients are real, the other root is $3+i$. The sum of the roots is $6=-p/2$, so $p=-12$. The product is $10=q/2$, so $q=20$. Thus $p+q=8$. The value -2 comes from using $q=10$ instead of $q=20$; 20 is just $q$; 32 comes from the sign error $p=12$ together with $q=20$.
2. What are the solutions of $x(x-4)+20=0$?
   Answer: $x=2 \pm 4i$ When you see a quadratic equation like $x(x-4)+20=0$, first expand and write it in standard form: $x^2-4x+20=0$. The discriminant is $(-4)^2-4(1)(20)=16-80=-64$, so the solutions will be complex. Applying the quadratic formula gives $$x=\frac{4\pm \sqrt{-64}}{2}=\frac{4\pm 8i}{2}=2\pm 4i.$$ That is why $x=2\pm 4i$ is correct: both the real part and the imaginary part are divided by $2$. The choice $x=4\pm 8i$ comes from finding $4\pm 8i$ but forgetting to divide by the $2$ in the quadratic formula. The choice $x=2\pm 8i$ correctly simplifies the real part but leaves the imaginary part unsimplified — $\frac{8i}{2}$ should become $4i$. The choice $x=-2\pm 4i$ has the correct imaginary part but the wrong sign on the real part; since the equation is $x^2-4x+20=0$, the quadratic formula uses $-(-4)=4$, and $\frac{4}{2}=2$, not $-2$. A reliable strategy is to complete the square as a check: $x^2-4x+20=0$ becomes $(x-2)^2+16=0$, so $(x-2)^2=-16$ and $x-2=\pm 4i$. This gives $x=2\pm 4i$ quickly and helps you catch sign or division errors. Whenever you see complex solutions, slow down and simplify both parts of the quadratic formula completely.
3. The roots of $x^2+4x+c=0$ are $r$ and $s$. If $r-s=6i$, what is $c$?
   Answer: 13 The roots are $-2\pm \sqrt{c-4}\,i$ for $c>4$, so their difference is $2\sqrt{c-4}\,i$. Setting $2\sqrt{c-4}\,i=6i$ gives $\sqrt{c-4}=3$, so $c=13$. The value -5 comes from treating the difference as real and solving $16-4c=36$; 7 comes from forgetting the square root in $2(c-4)=6$; 40 comes from dropping the factor 2 and solving $c-4=36$.
4. What are the solutions of the equation $x^2 + 6x + 25 = 0$?
   Answer: $x = -3 \pm 4i$ When you see a quadratic equation with a negative discriminant, expect complex conjugate solutions. The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ handles these directly. Here $a = 1$, $b = 6$, $c = 25$, so the discriminant is $6^2 - 4(1)(25) = 36 - 100 = -64$. Since $\sqrt{-64} = 8i$, the formula gives $$x = \frac{-6 \pm 8i}{2} = -3 \pm 4i.$$ That is the correct solution set. The choice $x = -6 \pm 8i$ appears if you forget to divide the entire numerator by $2a$ — only the $-6$ and the $8i$ each need to be halved. The choice $x = 3 \pm 4i$ has the wrong real-part sign; it comes from mistaking $-b/(2a)$ for positive $3$, but the vertex coordinate is $-3$. The choice $x = -3 \pm 8i$ correctly identifies the real part but forgets to divide the imaginary part by $2a$: $8i/2$ is $4i$, not $8i$. A strong check is completing the square: $x^2 + 6x + 25 = (x+3)^2 + 16 = 0$, so $(x+3)^2 = -16$, giving $x+3 = \pm 4i$, hence $x = -3 \pm 4i$. Whenever you use the quadratic formula with complex answers, remember the $\pm$ applies to the whole numerator and both parts must be divided by $2a$.
5. The discriminant of $ax^2+bx+c=0$, where $a,b,c$ are real and $a \ne 0$, is $-24$. Which statement must be true?
   Answer: There are two non-real complex roots, and they are conjugates. When you see a question about the discriminant of a quadratic, your first thought should be: what does the sign of the discriminant tell me about the roots? For $ax^2+bx+c=0$ with real coefficients, the discriminant $\Delta = b^2-4ac$ determines the nature of the solutions. Here $\Delta = -24$, which is negative. That means the quadratic has no real roots at all. Instead, it has two non-real complex roots. Moreover, because the coefficients $a,b,c$ are real, those two non-real roots must be complex conjugates of each other — for example, $p+qi$ and $p-qi$. So the statement "There are two non-real complex roots, and they are conjugates" is the one that must be true. Now look at the other choices. "There are no solutions in the complex number system" is false: every quadratic has two complex solutions counting multiplicity, by the Fundamental Theorem of Algebra. "There are two distinct real roots, and they are irrational" is impossible because a negative discriminant means no real roots at all. And "There is one real root and one non-real complex root" cannot happen for a quadratic with real coefficients — non-real roots always come in conjugate pairs, so you either have two real roots or two non-real roots, never one of each. Here's your takeaway: for a real-coefficient quadratic, positive discriminant means two distinct real roots, zero means one repeated real root, and negative means two non-real conjugate roots. The discriminant only tells you about irrational roots when it is positive and not a perfect square.
6. For $p\ne0$, which statement is true about $px^2+2px+2p=0$?
   Answer: Its roots are $-1\pm i$ for every nonzero $p$. Since $p\ne0$, divide the equation by $p$ to get $x^2+2x+2=0$. Its discriminant is $4-8=-4$, so the roots are $\frac{-2\pm 2i}{2}=-1\pm i$. The sign of $p$ does not change the equation after division, and the roots $-1\pm i\sqrt2$ would require discriminant $-8$.
7. Which quadratic equation has roots $-\frac12\pm \frac32 i$?
   Answer: $2x^2+2x+5=0$ The roots have sum $-1$ and product $\left(-\frac12\right)^2+\left(\frac32\right)^2=\frac14+\frac94=\frac52$. The monic equation is $x^2+x+\frac52=0$, and multiplying by 2 gives $2x^2+2x+5=0$. The other forms have a sign error in the linear term or a negative constant term, which changes the product of the roots.
8. A quadratic equation with real coefficients has $2-3i$ as one of its roots. Which of the following could be the equation?
   Answer: $x^2 - 4x + 13 = 0$ Whenever you see a quadratic with real coefficients and a complex root like $2-3i$, the first thing to remember is the conjugate root theorem: the other root must be $2+3i$. This is because complex roots of real-coefficient polynomials always come in conjugate pairs. So your two roots are $2-3i$ and $2+3i$. Their sum is $4$, and their product is $(2)^2 + (3)^2 = 13$. A quadratic with those roots is $x^2 - (\text{sum})x + (\text{product}) = x^2 - 4x + 13 = 0$, which is the correct equation. Now look at the wrong choices. $x^2 + 4x + 13 = 0$ has the right product $13$ but the wrong sum $-4$, so its roots would be $-2 \pm 3i$, not $2 \pm 3i$. The equations $x^2 - 4x - 13 = 0$ and $x^2 + 4x - 13 = 0$ both have a negative product, which means the two roots would have opposite signs — they cannot be a complex conjugate pair, whose product is always positive. These choices likely come from confusing the sign of the middle term or forgetting that the product of conjugates is $a^2+b^2$, not $a^2-b^2$. Your takeaway: for any real-coefficient quadratic, once you see one complex root, immediately write down its conjugate. Then build the equation from sum and product. Watch the signs carefully — the middle term is negative the sum, and the constant term is the product.
9. Let $f(x)=ax^2+bx+c$ have real coefficients, $a<0$, and $b^2-4ac=-3$. Which statement must be true?
   Answer: $f(x)<0$ for every real number $x$. When you see a quadratic with real coefficients and you're told $a<0$ and a discriminant, think about two things: what the discriminant says about real roots, and what the leading coefficient says about the parabola's shape. Here $b^2-4ac=-3<0$, so $f(x)=0$ has no real solutions: the graph never crosses the $x$-axis. Since $a<0$, the parabola opens downward. A downward-opening parabola that never touches the $x$-axis must lie entirely below that axis, so $f(x)<0$ for every real number $x$. The choice "exactly one real solution" would require the discriminant to equal zero, not negative. The choice "two distinct real solutions" would require a positive discriminant. The choice "graph of $f$ has a minimum value at its vertex" gets the direction exactly backward: with $a<0$, the vertex is the highest point on the graph, so it gives a maximum, not a minimum. A useful pattern to remember: if a quadratic has no real roots, its value never changes sign, and the sign is the same as the sign of $a$. Here that instantly shows $f(x)$ is always negative.
10. For which real values of $k$ does the equation $kx^2+2x+k=0$ have two distinct real roots?
    Answer: $-1<k<0$ or $0<k<1$ Whenever you see a question asking for real values of a parameter that make a quadratic have two distinct real roots, check two things: the leading coefficient must be nonzero, and the discriminant must be positive. Here $a=k$, $b=2$, and $c=k$, so $$b^2-4ac = 4-4k^2 = 4(1-k^2).$$ Requiring $4(1-k^2)>0$ gives $1-k^2>0$, so $-1<k<1$. But $k=0$ must be excluded because then the equation becomes $2x=0$, a linear equation with only one root. Therefore the valid values are $-1<k<0$ or $0<k<1$. The choice $-1<k<1$ is the trap of forgetting the $k\neq0$ condition: it includes $k=0$, which destroys the quadratic nature. The choice $k<-1$ or $k>1$ reverses the discriminant inequality: those $k$ values make $4(1-k^2)<0$, so the equation has no real roots, not two. The choice $0<k<1$ only considers positive $k$; it misses the symmetric negative interval, even though negative values like $k=-\frac12$ still produce a positive discriminant and two real roots. Study tip: whenever a quadratic has a parameter in the leading coefficient, always check that parameter separately from the discriminant. Write the discriminant inequality, then intersect it with the condition that the coefficient of $x^2$ is not zero.
11. Which statement is true for the equation $x^2-(k+2)x+(k^2+4)=0$ for every real number $k$?
    Answer: It has two non-real roots for every $k$. The discriminant is $D=(k+2)^2-4(k^2+4)=-3k^2+4k-12$. Completing the square gives $-3(k-\frac23)^2-\frac{32}{3}$, which is always negative. Therefore the equation has two non-real roots for every real $k$. There is no value of $k$ for which the discriminant is zero, so it never has exactly one real root.
12. The discriminant of $ax^2+bx+c=0$, where $a,b,c$ are real and $a \ne 0$, is $16$. Which statement must be true?
    Answer: There are two distinct real roots. When you see a discriminant for a quadratic with real coefficients, your first move is to check its sign — that tells you the type of roots before you ever solve the equation. Here the discriminant is $16$, which is positive. For real $a,b,c$, a positive discriminant means the quadratic has two distinct real roots. This is because the quadratic formula gives $$x=\frac{-b\pm \sqrt{16}}{2a}=\frac{-b\pm 4}{2a},$$ producing two different real values since $4\neq -4$. So "There are two distinct real roots" is the statement that must be true. Why are the others false? "There are two non-real complex roots" would require a negative discriminant, not a positive one. "There is exactly one real root" would require the discriminant to be $0$, giving a repeated root. And "There are two distinct rational roots" is tempting because $16$ is a perfect square, but that logic only works if $a,b,c$ are rational. Since the coefficients are only stated to be real, the two roots could be irrational — for example, $x^2+\sqrt{5}x-\frac{11}{4}=0$ has discriminant $16$ yet non-rational roots. Your takeaway: for quadratic roots, sign of the discriminant decides everything. Positive means two distinct real roots; zero means one real root; negative means two non-real complex roots. Don't assume rationality unless the coefficients are rational.
13. For which set of real values of $k$ does the equation $x^2 + (k-1)x + 1 = 0$ have two distinct non-real complex roots?
    Answer: $-1<k<3$ Whenever you see a question about the nature of the roots of a quadratic, your first move should be to examine the discriminant: for $ax^2+bx+c=0$, the discriminant is $D=b^2-4ac$. If $D<0$, the equation has two distinct non-real complex roots. Here $a=1$, $b=k-1$, and $c=1$, so $D=(k-1)^2-4(1)(1)=(k-1)^2-4.$ For non-real roots, require $D<0$: $(k-1)^2-4<0 \quad\Rightarrow\quad (k-1)^2<4.$ Taking square roots gives $|k-1|<2$, which means $-2<k-1<2 \quad\Rightarrow\quad -1<k<3.$ So the correct set is $-1<k<3$. Now look at the traps. The choice $k<-1$ or $k>3$ corresponds to $D>0$, which gives two distinct real roots — it is exactly the opposite of what is needed. The choice $k<3$ is too broad: it includes values like $k=-2$, where $D=9>0$, producing real roots. Similarly, $k>-1$ is too broad: it includes values like $k=4$, where $D=9>0$, again producing real roots. Both one-sided choices miss the need for the full interval between $-1$ and $3$. Study tip: for quadratic root-type questions, always set the discriminant sign correctly — negative for two non-real complex roots, positive for two real roots, zero for one repeated real root — then solve the inequality carefully, including both bounds.
14. A student solves $x^2-4x+6=0$ and reports $x=2\pm 2i\sqrt2$. Which statement best identifies the error?
    Answer: The student simplified $\frac{4\pm 2i\sqrt2}{2}$ incorrectly, dividing only the first term by 2. The discriminant is $16-24=-8$, so the quadratic formula gives $x=\frac{4\pm 2i\sqrt2}{2}=2\pm i\sqrt2$. The student's answer $2\pm 2i\sqrt2$ is obtained by dividing only the $4$ by 2 and leaving the radical term undivided. The other suggested errors would not produce the reported expression.
15. A quadratic function $f$ has vertex $(3,-5)$ and opens downward. Which statement must be true about the solutions of $f(x)=0$?
    Answer: There are no real solutions; the solutions are a nonreal conjugate pair. Opening downward means the vertex is the maximum. Since the maximum value is $-5<0$, the entire parabola lies below the x-axis, so it has no x-intercepts. A quadratic with real coefficients then has two nonreal conjugate roots. Two real solutions on opposite sides of 3 would be the behavior of an upward-opening parabola with vertex below the x-axis, not this function.
16. Let $r$ and $s$ be the roots of $x^2-4x+13=0$. What is $r^2+s^2$?
    Answer: $-10$ When you see a question about the roots of a quadratic, Vieta's formulas are your fastest tool. For $x^2-4x+13=0$, the sum of the roots is $$r+s=4,$$ and the product is $$rs=13.$$ You want $r^2+s^2$, so rewrite it using the square of the sum: $$r^2+s^2=(r+s)^2-2rs.$$ Substitute the values: $$r^2+s^2=4^2-2(13)=16-26=-10.$$ So the correct value is $-10$. Now let's look at the traps. Choosing $10$ often comes from misreading the constant term $13$ as $3$, then computing $16-2(3)=10$. The $42$ choice comes from a sign error in the formula: adding $2rs$ instead of subtracting it, giving $16+26=42$. The $-42$ choice may come from computing $42$ and then forcing a negative sign because the roots are non-real; however, complex roots do not automatically make every expression negative. You must rely on the algebra, not on the discriminant's sign. Remember: for any quadratic, express symmetric root expressions like $r^2+s^2$, $\frac1r+\frac1s$, or $r^3+s^3$ in terms of $r+s$ and $rs$. Vieta's formulas work for complex roots too, so don't be thrown off by a negative discriminant.
17. For which real values of $k$ does $kx^2+3x+k=0$ have two non-real roots?
    Answer: $k<-\frac32$ or $k>\frac32$ For $k\ne0$, the discriminant is $9-4k^2$. Two non-real roots require $9-4k^2<0$, so $k^2>\frac94$, which means $k<-\frac32$ or $k>\frac32$. If $k=0$, the equation is linear and has only one real root. The open interval between $-\frac32$ and $\frac32$ gives two distinct real roots, and including equality gives one repeated real root.
18. For which real value(s) of $k$ does the equation $x^2+kx+k=0$ have exactly one distinct real solution?
    Answer: $k=0$ or $k=4$ Whenever you see a quadratic equation and are asked when it has exactly one distinct real solution, think discriminant. For $ax^2+bx+c=0$, the condition is $b^2-4ac=0$. Here $a=1$, $b=k$, and $c=k$, so set $k^2-4k=0$. Factoring gives $k(k-4)=0$, so $k=0$ or $k=4$. Both values truly work: with $k=0$, the equation is $x^2=0$, and with $k=4$, it becomes $x^2+4x+4=(x+2)^2=0$. Each has exactly one distinct real root. Now look at the other choices. The option "$k=0$ or $k=-4$" likely comes from writing the discriminant as $k^2+4k$ instead of $k^2-4k$. If $k=-4$, the equation is $x^2-4x-4=0$, whose discriminant is $16+16=32>0$, giving two real solutions. The choice "$k=4$ only" overlooks $k=0$, but $x^2=0$ is a perfect square with one distinct root, so $k=0$ is valid. The choice "$k=0$ only" overlooks $k=4$, even though $x^2+4x+4=(x+2)^2$ clearly has one distinct root. Strategy: for any quadratic parameter question about "exactly one real solution" or a double root, set the discriminant equal to zero first. Then verify each solution by plugging it back into the original equation or factoring, so you don't accidentally drop a valid value like $k=0$.
19. For how many real values of $k$ does $(k-2)x^2+(2k-4)x+(k-2)=0$ have two non-real roots?
    Answer: 0 If $k=2$, the equation becomes $0=0$, which is not a quadratic and has infinitely many real solutions, not two non-real roots. If $k\ne2$, factor as $(k-2)(x+1)^2=0$, so the only solution is $x=-1$, a repeated real root. Equivalently, the discriminant is $(2k-4)^2-4(k-2)^2=0$ for every $k$, so non-real roots never occur.
20. Which equation has a leading coefficient of 2 and one root $-1+i\sqrt2$?
    Answer: $2x^2+4x+6=0$ The conjugate root is $-1-i\sqrt2$, so the sum is $-2$ and the product is $1+2=3$. With leading coefficient 2, the equation is $2(x^2+2x+3)=2x^2+4x+6=0$. The option with $-4x+6$ has roots $1\pm i\sqrt2$; the options with $-6$ have real roots because their product term is negative.

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