Algebra 3 Quiz: Complex Roots Of Quadratics
12 questions · exam conditions
0:00
Complex Roots Of QuadraticsQuestion 1 of 12

What are the solutions of the equation x2+6x+25=0x^2 + 6x + 25 = 0?

x=−3±4ix = -3 \pm 4i
x=−6±8ix = -6 \pm 8i
x=3±4ix = 3 \pm 4i
x=−3±8ix = -3 \pm 8i
← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Complex Roots Of Quadratics

Practice Complex Roots Of Quadratics in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complex Roots Of Quadratics, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What are the solutions of the equation x2+6x+25=0x^2 + 6x + 25 = 0?

  1. x=−3±4ix = -3 \pm 4i (correct answer)
  2. x=−6±8ix = -6 \pm 8i
  3. x=3±4ix = 3 \pm 4i
  4. x=−3±8ix = -3 \pm 8i
Explanation: When you see a quadratic equation with a negative discriminant, expect complex conjugate solutions. The quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} handles these directly. Here a=1a = 1, b=6b = 6, c=25c = 25, so the discriminant is 62−4(1)(25)=36−100=−646^2 - 4(1)(25) = 36 - 100 = -64. Since −64=8i\sqrt{-64} = 8i, the formula gives x=−6±8i2=−3±4i.x = \frac{-6 \pm 8i}{2} = -3 \pm 4i. That is the correct solution set. The choice x=−6±8ix = -6 \pm 8i appears if you forget to divide the entire numerator by 2a2a — only the −6-6 and the 8i8i each need to be halved. The choice x=3±4ix = 3 \pm 4i has the wrong real-part sign; it comes from mistaking −b/(2a)-b/(2a) for positive 33, but the vertex coordinate is −3-3. The choice x=−3±8ix = -3 \pm 8i correctly identifies the real part but forgets to divide the imaginary part by 2a2a: 8i/28i/2 is 4i4i, not 8i8i. A strong check is completing the square: x2+6x+25=(x+3)2+16=0x^2 + 6x + 25 = (x+3)^2 + 16 = 0, so (x+3)2=−16(x+3)^2 = -16, giving x+3=±4ix+3 = \pm 4i, hence x=−3±4ix = -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 2a2a.

Question 2

For which real value(s) of kk does the equation x2+kx+k=0x^2+kx+k=0 have exactly one distinct real solution?

  1. k=0k=0 or k=−4k=-4
  2. k=4k=4 only
  3. k=0k=0 only
  4. k=0k=0 or k=4k=4 (correct answer)
Explanation: Whenever you see a quadratic equation and are asked when it has exactly one distinct real solution, think discriminant. For ax2+bx+c=0ax^2+bx+c=0, the condition is b2−4ac=0b^2-4ac=0. Here a=1a=1, b=kb=k, and c=kc=k, so set k2−4k=0k^2-4k=0. Factoring gives k(k−4)=0k(k-4)=0, so k=0k=0 or k=4k=4. Both values truly work: with k=0k=0, the equation is x2=0x^2=0, and with k=4k=4, it becomes x2+4x+4=(x+2)2=0x^2+4x+4=(x+2)^2=0. Each has exactly one distinct real root. Now look at the other choices. The option "k=0k=0 or k=−4k=-4" likely comes from writing the discriminant as k2+4kk^2+4k instead of k2−4kk^2-4k. If k=−4k=-4, the equation is x2−4x−4=0x^2-4x-4=0, whose discriminant is 16+16=32>016+16=32>0, giving two real solutions. The choice "k=4k=4 only" overlooks k=0k=0, but x2=0x^2=0 is a perfect square with one distinct root, so k=0k=0 is valid. The choice "k=0k=0 only" overlooks k=4k=4, even though x2+4x+4=(x+2)2x^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=0k=0.

Question 3

One solution of x2+bx+c=0x^2+bx+c=0 is 3+2i3+2i. If bb and cc are real, what are bb and cc?

  1. b=−6b=-6, c=−13c=-13
  2. b=6b=6, c=13c=13
  3. b=−6b=-6, c=5c=5
  4. b=−6b=-6, c=13c=13 (correct answer)
Explanation: When you see a quadratic with real coefficients and one complex root, remember that complex roots must come in conjugate pairs. Since 3+2i3+2i is a root, 3−2i3-2i must also be a root. In x2+bx+c=0x^2+bx+c=0, the sum of the roots equals −b-b, and the product of the roots equals cc. Adding the conjugates gives (3+2i)+(3−2i)=6(3+2i)+(3-2i)=6, so −b=6-b=6, meaning b=−6b=-6. Multiplying gives (3+2i)(3−2i)=9−4i2=9+4=13(3+2i)(3-2i)=9-4i^2=9+4=13, so c=13c=13. Therefore, the correct pair is b=−6b=-6, c=13c=13. The choice b=−6b=-6, c=−13c=-13 has the correct bb, but it incorrectly treats the product as negative instead of recognizing that multiplying conjugates gives a positive real number. The choice b=6b=6, c=13c=13 correctly finds cc, but it takes the sum of the roots as bb rather than as −b-b. The choice b=−6b=-6, c=5c=5 also has the correct bb, but it computes the product as 9−49-4, forgetting that i2=−1i^2=-1; the product is actually 9+4=139+4=13. For any quadratic, remember: sum of roots is −ba-\frac{b}{a} and product is ca\frac{c}{a}. When coefficients are real, a complex root always brings its conjugate along.

Question 4

The roots of a quadratic equation with leading coefficient 1 are a+bia+bi and a−bia-bi, where aa and bb are real and b≠0b \ne 0. Which of the following is the equation?

  1. x2−2ax+(a2−b2)=0x^2-2ax+(a^2-b^2)=0
  2. x2−2ax+(a2+b2)=0x^2-2ax+(a^2+b^2)=0 (correct answer)
  3. x2+2ax+(a2+b2)=0x^2+2ax+(a^2+b^2)=0
  4. x2+2ax+(a2−b2)=0x^2+2ax+(a^2-b^2)=0
Explanation: Whenever you see roots like a+bia+bi and a−bia-bi, recall that for any quadratic with leading coefficient 11, the equation is x2−(sum of roots)x+(product of roots)=0.x^2-(\text{sum of roots})x+(\text{product of roots})=0. So your task is just to compute the sum and product of the two roots. The sum is (a+bi)+(a−bi)=2a,(a+bi)+(a-bi)=2a, so the xx-coefficient must be −2a-2a. The product is (a+bi)(a−bi)=a2−(bi)2=a2−b2i2.(a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2. Since i2=−1i^2=-1, this becomes a2−b2(−1)=a2+b2.a^2-b^2(-1)=a^2+b^2. Therefore the equation is x2−2ax+(a2+b2)=0.x^2-2ax+(a^2+b^2)=0. The choice with x2−2ax+(a2−b2)x^2-2ax+(a^2-b^2) uses the wrong product: it forgets that (bi)2=−b2(bi)^2=-b^2, not b2b^2. The choice with x2+2ax+(a2+b2)x^2+2ax+(a^2+b^2) has the right product but the wrong sign on the xx-term, because the sum is 2a2a, so the coefficient must be −2a-2a. The choice with x2+2ax+(a2−b2)x^2+2ax+(a^2-b^2) makes both errors, reversing the sum sign and mishandling i2i^2. A reliable study takeaway: for any quadratic with leading coefficient 11, immediately write x2−(sum)x+(product)=0.x^2-(\text{sum})x+(\text{product})=0. For complex conjugate roots, the sum is always real (2a2a), and the product is always positive (a2+b2a^2+b^2). If you remember that i2=−1i^2=-1, you will avoid the most common trap in this problem.

Question 5

For which set of real values of kk does the equation x2+(k−1)x+1=0x^2 + (k-1)x + 1 = 0 have two distinct non-real complex roots?

  1. −1<k<3-1<k<3 (correct answer)
  2. k<−1k<-1 or k>3k>3
  3. k<3k<3
  4. k>−1k>-1
Explanation: Whenever you see a question about the nature of the roots of a quadratic, your first move should be to examine the discriminant: for ax2+bx+c=0ax^2+bx+c=0, the discriminant is D=b2−4acD=b^2-4ac. If D<0D<0, the equation has two distinct non-real complex roots. Here a=1a=1, b=k−1b=k-1, and c=1c=1, so D=(k−1)2−4(1)(1)=(k−1)2−4.D=(k-1)^2-4(1)(1)=(k-1)^2-4. For non-real roots, require D<0D<0: (k−1)2−4<0⇒(k−1)2<4.(k-1)^2-4<0 \quad\Rightarrow\quad (k-1)^2<4. Taking square roots gives ∣k−1∣<2|k-1|<2, which means −2<k−1<2⇒−1<k<3.-2<k-1<2 \quad\Rightarrow\quad -1<k<3. So the correct set is −1<k<3-1<k<3. Now look at the traps. The choice k<−1k<-1 or k>3k>3 corresponds to D>0D>0, which gives two distinct real roots — it is exactly the opposite of what is needed. The choice k<3k<3 is too broad: it includes values like k=−2k=-2, where D=9>0D=9>0, producing real roots. Similarly, k>−1k>-1 is too broad: it includes values like k=4k=4, where D=9>0D=9>0, again producing real roots. Both one-sided choices miss the need for the full interval between −1-1 and 33. 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.

Question 6

What are the solutions of x(x−4)+20=0x(x-4)+20=0?

  1. x=4±8ix=4 \pm 8i
  2. x=2±8ix=2 \pm 8i
  3. x=2±4ix=2 \pm 4i (correct answer)
  4. x=−2±4ix=-2 \pm 4i
Explanation: When you see a quadratic equation like x(x−4)+20=0x(x-4)+20=0, first expand and write it in standard form: x2−4x+20=0x^2-4x+20=0. The discriminant is (−4)2−4(1)(20)=16−80=−64(-4)^2-4(1)(20)=16-80=-64, so the solutions will be complex. Applying the quadratic formula gives x=4±−642=4±8i2=2±4i.x=\frac{4\pm \sqrt{-64}}{2}=\frac{4\pm 8i}{2}=2\pm 4i. That is why x=2±4ix=2\pm 4i is correct: both the real part and the imaginary part are divided by 22. The choice x=4±8ix=4\pm 8i comes from finding 4±8i4\pm 8i but forgetting to divide by the 22 in the quadratic formula. The choice x=2±8ix=2\pm 8i correctly simplifies the real part but leaves the imaginary part unsimplified — 8i2\frac{8i}{2} should become 4i4i. The choice x=−2±4ix=-2\pm 4i has the correct imaginary part but the wrong sign on the real part; since the equation is x2−4x+20=0x^2-4x+20=0, the quadratic formula uses −(−4)=4-(-4)=4, and 42=2\frac{4}{2}=2, not −2-2. A reliable strategy is to complete the square as a check: x2−4x+20=0x^2-4x+20=0 becomes (x−2)2+16=0(x-2)^2+16=0, so (x−2)2=−16(x-2)^2=-16 and x−2=±4ix-2=\pm 4i. This gives x=2±4ix=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.

Question 7

Let rr and ss be the roots of x2−4x+13=0x^2-4x+13=0. What is r2+s2r^2+s^2?

  1. 1010
  2. 4242
  3. −10-10 (correct answer)
  4. −42-42
Explanation: When you see a question about the roots of a quadratic, Vieta's formulas are your fastest tool. For x2−4x+13=0x^2-4x+13=0, the sum of the roots is r+s=4,r+s=4, and the product is rs=13.rs=13. You want r2+s2r^2+s^2, so rewrite it using the square of the sum: r2+s2=(r+s)2−2rs.r^2+s^2=(r+s)^2-2rs. Substitute the values: r2+s2=42−2(13)=16−26=−10.r^2+s^2=4^2-2(13)=16-26=-10. So the correct value is −10-10. Now let's look at the traps. Choosing 1010 often comes from misreading the constant term 1313 as 33, then computing 16−2(3)=1016-2(3)=10. The 4242 choice comes from a sign error in the formula: adding 2rs2rs instead of subtracting it, giving 16+26=4216+26=42. The −42-42 choice may come from computing 4242 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 r2+s2r^2+s^2, 1r+1s\frac1r+\frac1s, or r3+s3r^3+s^3 in terms of r+sr+s and rsrs. Vieta's formulas work for complex roots too, so don't be thrown off by a negative discriminant.

Question 8

The discriminant of ax2+bx+c=0ax^2+bx+c=0, where a,b,ca,b,c are real and a≠0a \ne 0, is −24-24. Which statement must be true?

  1. There are no solutions in the complex number system.
  2. There are two distinct real roots, and they are irrational.
  3. There is one real root and one non-real complex root.
  4. There are two non-real complex roots, and they are conjugates. (correct answer)
Explanation: 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 ax2+bx+c=0ax^2+bx+c=0 with real coefficients, the discriminant Δ=b2−4ac\Delta = b^2-4ac determines the nature of the solutions. Here Δ=−24\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,ca,b,c are real, those two non-real roots must be complex conjugates of each other — for example, p+qip+qi and p−qip-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.

Question 9

A quadratic equation with real coefficients has 2−3i2-3i as one of its roots. Which of the following could be the equation?

  1. x2−4x+13=0x^2 - 4x + 13 = 0 (correct answer)
  2. x2+4x+13=0x^2 + 4x + 13 = 0
  3. x2−4x−13=0x^2 - 4x - 13 = 0
  4. x2+4x−13=0x^2 + 4x - 13 = 0
Explanation: Whenever you see a quadratic with real coefficients and a complex root like 2−3i2-3i, the first thing to remember is the conjugate root theorem: the other root must be 2+3i2+3i. This is because complex roots of real-coefficient polynomials always come in conjugate pairs. So your two roots are 2−3i2-3i and 2+3i2+3i. Their sum is 44, and their product is (2)2+(3)2=13(2)^2 + (3)^2 = 13. A quadratic with those roots is x2−(sum)x+(product)=x2−4x+13=0x^2 - (\text{sum})x + (\text{product}) = x^2 - 4x + 13 = 0, which is the correct equation. Now look at the wrong choices. x2+4x+13=0x^2 + 4x + 13 = 0 has the right product 1313 but the wrong sum −4-4, so its roots would be −2±3i-2 \pm 3i, not 2±3i2 \pm 3i. The equations x2−4x−13=0x^2 - 4x - 13 = 0 and x2+4x−13=0x^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 a2+b2a^2+b^2, not a2−b2a^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.

Question 10

For which real values of kk does the equation kx2+2x+k=0kx^2+2x+k=0 have two distinct real roots?

  1. −1<k<1-1<k<1
  2. −1<k<0-1<k<0 or 0<k<10<k<1 (correct answer)
  3. k<−1k<-1 or k>1k>1
  4. 0<k<10<k<1
Explanation: 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=ka=k, b=2b=2, and c=kc=k, so b2−4ac=4−4k2=4(1−k2).b^2-4ac = 4-4k^2 = 4(1-k^2). Requiring 4(1−k2)>04(1-k^2)>0 gives 1−k2>01-k^2>0, so −1<k<1-1<k<1. But k=0k=0 must be excluded because then the equation becomes 2x=02x=0, a linear equation with only one root. Therefore the valid values are −1<k<0-1<k<0 or 0<k<10<k<1. The choice −1<k<1-1<k<1 is the trap of forgetting the k≠0k\neq0 condition: it includes k=0k=0, which destroys the quadratic nature. The choice k<−1k<-1 or k>1k>1 reverses the discriminant inequality: those kk values make 4(1−k2)<04(1-k^2)<0, so the equation has no real roots, not two. The choice 0<k<10<k<1 only considers positive kk; it misses the symmetric negative interval, even though negative values like k=−12k=-\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 x2x^2 is not zero.

Question 11

The discriminant of ax2+bx+c=0ax^2+bx+c=0, where a,b,ca,b,c are real and a≠0a \ne 0, is 1616. Which statement must be true?

  1. There are two non-real complex roots.
  2. There are two distinct rational roots.
  3. There are two distinct real roots. (correct answer)
  4. There is exactly one real root.
Explanation: 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 1616, which is positive. For real a,b,ca,b,c, a positive discriminant means the quadratic has two distinct real roots. This is because the quadratic formula gives x=−b±162a=−b±42a,x=\frac{-b\pm \sqrt{16}}{2a}=\frac{-b\pm 4}{2a}, producing two different real values since 4≠−44\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 00, giving a repeated root. And "There are two distinct rational roots" is tempting because 1616 is a perfect square, but that logic only works if a,b,ca,b,c are rational. Since the coefficients are only stated to be real, the two roots could be irrational — for example, x2+5x−114=0x^2+\sqrt{5}x-\frac{11}{4}=0 has discriminant 1616 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.

Question 12

Let f(x)=ax2+bx+cf(x)=ax^2+bx+c have real coefficients, a<0a<0, and b2−4ac=−3b^2-4ac=-3. Which statement must be true?

  1. The equation f(x)=0f(x)=0 has exactly one real solution.
  2. f(x)<0f(x)<0 for every real number xx. (correct answer)
  3. The equation f(x)=0f(x)=0 has two distinct real solutions.
  4. The graph of ff has a minimum value at its vertex.
Explanation: When you see a quadratic with real coefficients and you're told a<0a<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 b2−4ac=−3<0b^2-4ac=-3<0, so f(x)=0f(x)=0 has no real solutions: the graph never crosses the xx-axis. Since a<0a<0, the parabola opens downward. A downward-opening parabola that never touches the xx-axis must lie entirely below that axis, so f(x)<0f(x)<0 for every real number xx. 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 ff has a minimum value at its vertex" gets the direction exactly backward: with a<0a<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 aa. Here that instantly shows f(x)f(x) is always negative.