Algebra Quiz: Solving Quadratic Equations With Complex Solutions

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Solving Quadratic Equations With Complex SolutionsQuestion 1 of 20

Solve using the quadratic formula: $2x^2 + 4x + 5 = 0$ . Express solutions in $a + bi$ form (use $i^2 = -1$ ).

x = -1 \pm \tfrac{\sqrt{6}}{2}i

x = -1 \pm \tfrac{\sqrt{6}}{4}i

x = 1 \pm \tfrac{\sqrt{6}}{2}i

x = -1 \pm \tfrac{\sqrt{6}}{2}

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Algebra Quiz: Solving Quadratic Equations With Complex Solutions

Practice Solving Quadratic Equations With Complex Solutions in Algebra 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 Solving Quadratic Equations With Complex Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

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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.

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Question 1

Solve using the quadratic formula: $2x^2 + 4x + 5 = 0$ . Express solutions in $a + bi$ form (use $i^2 = -1$ ).

$x = -1 \pm \tfrac{\sqrt{6}}{2}i$ (correct answer)
$x = -1 \pm \tfrac{\sqrt{6}}{4}i$
$x = 1 \pm \tfrac{\sqrt{6}}{2}i$
$x = -1 \pm \tfrac{\sqrt{6}}{2}$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. When a quadratic equation has discriminant b² - 4ac < 0 (negative), the quadratic formula gives √(negative), which means complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). We separate the real part (-b/(2a)) from the imaginary part (±√|b² - 4ac|/(2a))i to write in a + bi form. The solutions always come in conjugate pairs: a + bi and a - bi. Solving 2x² + 4x + 5 = 0 using the quadratic formula: (1) Identify a = 2, b = 4, c = 5. (2) Calculate discriminant: b² - 4ac = (4)² - 4(2)(5) = 16 - 40 = -24. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-4 ± √(-24))/(2·2) = (-4 ± i√24)/4. (5) Simplify: √24 = √(4·6) = 2√6, so x = (-4 ± 2i√6)/4 = -4/4 ± 2i√6/4 = -1 ± (√6/2)i. Solutions: x = -1 + (√6/2)i and x = -1 - (√6/2)i. Choice A correctly solves to get x = -1 ± (√6/2)i with accurate calculation using i² = -1. Choice B has the wrong imaginary coefficient: from the quadratic formula with 2a = 4 in the denominator, we get 2i√6/4 = (√6/2)i, not (√6/4)i. The entire numerator (-4 ± 2i√6) must be divided by 4, giving -1 ± (√6/2)i. Always divide both parts by 2a! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -20 and you end up with no i in your answer, something went wrong.

Question 2

Solve using the quadratic formula: $x^2+2x+2=0$ . Write solutions in $a+bi$ form (use $i^2=-1$ ).

$x=-2\pm i$
$x=-1\pm 2i$
$x=-1\pm i$ (correct answer)
$x=1\pm i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. The imaginary unit i is defined as i = √(-1), which means i² = -1: this allows us to take square roots of negative numbers. When we have √(-16), we can write it as i√16 = 4i. Complex numbers have the form a + bi where a is the real part and b is the imaginary part (like 3 + 2i or -1 - 4i). These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving x² + 2x + 2 = 0 using the quadratic formula: (1) Identify a = 1, b = 2, c = 2. (2) Calculate discriminant: 2² - 4(1)(2) = 4 - 8 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-2 ± √(-4))/2 = (-2 ± i√4)/2. (5) Simplify: x = -2/2 ± (2i)/2 = -1 ± i. Solutions: x = -1 + i and x = -1 - i. Choice A correctly solves to get x = -1 ± i with accurate calculation using i² = -1. Choice C has the real and imaginary parts switched or calculated wrong: the imaginary part is 1, not 2; from the quadratic formula x = (-b ± i√|discriminant|)/(2a), the real part is -b/(2a) = -1, and the imaginary part is √|discriminant|/(2a) = 1; keep real and imaginary separate in your calculation! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -20 and you end up with no i in your answer, something went wrong. Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!

Question 3

Solve using the quadratic formula: $x^2-4x+5=0$ . Express solutions in $a+bi$ form, using $i^2=-1$ .

$x=2\pm i$ (correct answer)
$x=-2\pm i$
$x=2\pm 2i$
$x=2\pm 1$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. When a quadratic equation has discriminant b² - 4ac < 0 (negative), the quadratic formula gives √(negative), which means complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). We separate the real part (-b/(2a)) from the imaginary part (±√|b² - 4ac|/(2a))i to write in a + bi form. The solutions always come in conjugate pairs: a + bi and a - bi. Solving x² - 4x + 5 = 0 using the quadratic formula: (1) Identify a = 1, b = -4, c = 5. (2) Calculate discriminant: b² - 4ac = (-4)² - 4(1)(5) = 16 - 20 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (4 ± √(-4))/2 = (4 ± 2i)/2. (5) Simplify: x = 4/2 ± (2/2)i = 2 ± i. Solutions: x = 2 + i and x = 2 - i. Choice A correctly solves to get x = 2 ± i with accurate calculation using i² = -1. Choice D gives real solutions when the discriminant is negative. Check: b² - 4ac = 16 - 20 = -4. Negative discriminant means NO real solutions—only complex ones. When you get √(negative) in the formula, you can't ignore it or treat it as positive. You must include i! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -4 and you end up with no i in your answer, something went wrong.

Question 4

Solve $x^2+8x+20=0$ . Express answers as complex numbers in $a\pm bi$ form, using $i^2=-1$ .

$x=-4\pm 2i$ (correct answer)
$x=4\pm 2i$
$x=-4\pm i$
$x=-4\pm 2$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 8x + 20 = 0 using the quadratic formula: (1) Identify a = 1, b = 8, c = 20. (2) Calculate discriminant: b² - 4ac = (8)² - 4(1)(20) = 64 - 80 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-8 ± √(-16))/2 = (-8 ± 4i)/2. (5) Simplify: x = -8/2 ± (4/2)i = -4 ± 2i. Solutions: x = -4 + 2i and x = -4 - 2i. Choice A correctly solves to get x = -4 ± 2i with accurate calculation using i² = -1. Choice D gives real solutions when the discriminant is negative. Check: b² - 4ac = 64 - 80 = -16. Negative discriminant means NO real solutions—only complex ones. When you get √(negative) in the formula, you can't ignore it or treat it as positive. You must include i! Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!

Question 5

Complete the square to solve $x^2-4x+5=0$ for complex solutions. Use $i^2=-1$ and write the answers in $a\pm bi$ form.

$x=2\pm 2i$
$x=2\pm i$ (correct answer)
$x=-2\pm i$
$x=2\pm 1$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. When a quadratic equation has discriminant b² - 4ac < 0 (negative), the quadratic formula gives √(negative), which means complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). We separate the real part (-b/(2a)) from the imaginary part (±√|b² - 4ac|/(2a))i to write in a + bi form. The solutions always come in conjugate pairs: a + bi and a - bi. Solving x² - 4x + 5 = 0 by completing the square: (1) Move constant: x² - 4x = -5. (2) Add (b/2)²: x² - 4x + 4 = -5 + 4 = -1. (3) Factor: (x - 2)² = -1. (4) Take square roots: x - 2 = ±√(-1) = ±i. (5) Solve: x = 2 ± i. The negative value after completing the square signals complex solutions! Choice B correctly identifies x = 2 ± i with accurate calculation using i² = -1. Choice D gives real solutions when the discriminant is negative. Check: b² - 4ac = (-4)² - 4(1)(5) = 16 - 20 = -4. Negative discriminant means NO real solutions—only complex ones. When you get √(negative) in the formula, you can't ignore it or treat it as positive. You must include i! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -4 and you end up with no i in your answer, something went wrong.

Question 6

Solve $x^2-6x+13=0$ by completing the square. Use the imaginary unit $i$ (where $i^2=-1$ ) and express solutions in $a\pm bi$ form.

$x=3\pm 2i$ (correct answer)
$x=-3\pm 2i$
$x=3\pm 4i$
$x=6\pm 2i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. Complex solutions appear in conjugate pairs for quadratics with real coefficients: if 2 + 3i is a solution, then 2 - 3i is automatically also a solution (same real part, opposite imaginary part). This pairing is guaranteed by the ± in the quadratic formula and has important implications: the sum is real (4 + 6i + 4 - 6i = 8), and the product is real ((2+3i)(2-3i) = 4 + 9 = 13), which is why quadratics with complex roots can still have real coefficients! Solving x² - 6x + 13 = 0 by completing the square: (1) Move constant: x² - 6x = -13. (2) Add (b/2)²: x² - 6x + 9 = -13 + 9 = -4. (3) Factor: (x - 3)² = -4. (4) Take square roots: x - 3 = ±√(-4) = ±i√4 = ±2i. (5) Solve: x = 3 ± 2i. The negative value after completing the square signals complex solutions! Choice A correctly identifies x = 3 ± 2i with accurate calculation using i² = -1. Choice B makes a sign error: the real part should be +3, not -3. With the ± and the i both appearing in solutions, it's easy to lose track of signs. Write out each solution separately: x = 3 + 2i AND x = 3 - 2i to make sure both signs are correct. Simplifying with i: remember that i² = -1 is the key rule. When you expand (a + bi)², you get a² + 2abi + (bi)² = a² + 2abi + b²i² = a² + 2abi - b² (since i² = -1). Combine reals and imaginaries separately: (a² - b²) + (2ab)i. This is how verification works—expand, use i² = -1, group real and imaginary parts.

Question 7

Solve $x^2 + 8x + 20 = 0$ and write solutions in the form $a \pm bi$ . (Use $i^2 = -1$ .)

$x = -4 \pm 2$
$x = -4 \pm i$
$x = 4 \pm 2i$
$x = -4 \pm 2i$ (correct answer)

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. The imaginary unit i is defined as i = √(-1), which means i² = -1: this allows us to take square roots of negative numbers. When we have √(-16), we can write it as i√16 = 4i. Complex numbers have the form a + bi where a is the real part and b is the imaginary part (like 3 + 2i or -1 - 4i). These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving x² + 8x + 20 = 0 using the quadratic formula: (1) Identify a = 1, b = 8, c = 20. (2) Calculate discriminant: b² - 4ac = (8)² - 4(1)(20) = 64 - 80 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-8 ± √(-16))/2 = (-8 ± 4i)/2. (5) Simplify: x = -8/2 ± 4i/2 = -4 ± 2i. Solutions: x = -4 + 2i and x = -4 - 2i. Choice A correctly solves to get x = -4 ± 2i with accurate calculation using i² = -1. Choice D forgets the i in the solution: when the discriminant is negative, we have √(negative) = i√(positive), so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get √(negative number), write it as i√(positive number): √(-16) = i√16 = 4i, √(-5) = i√5, (3) Simplify the entire expression to a + bi form by separating real parts from i-parts, (4) Write both solutions using ±: x = a ± bi means x = a + bi and x = a - bi. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative.

Question 8

Solve using the quadratic formula: $x^2 + 2x + 5 = 0$ . Since the discriminant is negative, the solutions are complex. Use $i^2=-1$ and express both solutions in $a+bi$ form.

$x=-1\pm 2i$ (correct answer)
$x=1\pm 2i$
$x=-1\pm 4i$
$x=-2\pm 2i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. The imaginary unit i is defined as i = √(-1), which means i² = -1: this allows us to take square roots of negative numbers. When we have √(-16), we can write it as i√16 = 4i. Complex numbers have the form a + bi where a is the real part and b is the imaginary part (like 3 + 2i or -1 - 4i). These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving x² + 2x + 5 = 0 using the quadratic formula: (1) Identify a = 1, b = 2, c = 5. (2) Calculate discriminant: b² - 4ac = (2)² - 4(1)(5) = 4 - 20 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-2 ± √(-16))/2 = (-2 ± 4i)/2. (5) Simplify: x = -2/2 ± 4i/2 = -1 ± 2i. Solutions: x = -1 + 2i and x = -1 - 2i. Choice A correctly solves to get x = -1 ± 2i with accurate calculation using i² = -1. Choice B has the wrong real part: from the quadratic formula x = (-b ± i√|discriminant|)/(2a), the real part is -b/(2a) = -2/2 = -1, not 1, and the imaginary part is √16/2 = 4/2 = 2. Keep real and imaginary separate in your calculation! Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get √(negative number), write it as i√(positive number): √(-16) = i√16 = 4i, √(-5) = i√5, (3) Simplify the entire expression to a + bi form by separating real parts from i-parts, (4) Write both solutions using ±: x = a ± bi means x = a + bi and x = a - bi. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative.

Question 9

Solve $x^2 - 6x + 10 = 0$ . Express solutions in $a + bi$ form, using $i^2 = -1$ .

$x = 3 \pm 1$
$x = -3 \pm i$
$x = 3 \pm i$ (correct answer)
$x = 3 \pm 2i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. Complex solutions appear in conjugate pairs for quadratics with real coefficients: if 2 + 3i is a solution, then 2 - 3i is automatically also a solution (same real part, opposite imaginary part). This pairing is guaranteed by the ± in the quadratic formula and has important implications: the sum is real (4 + 6i + 4 - 6i = 8), and the product is real ((2+3i)(2-3i) = 4 + 9 = 13), which is why quadratics with complex roots can still have real coefficients! Solving x² - 6x + 10 = 0 using the quadratic formula: (1) Identify a = 1, b = -6, c = 10. (2) Calculate discriminant: b² - 4ac = (-6)² - 4(1)(10) = 36 - 40 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (6 ± √(-4))/2 = (6 ± 2i)/2. (5) Simplify: x = 6/2 ± 2i/2 = 3 ± i. Solutions: x = 3 + i and x = 3 - i. Choice A correctly solves to get x = 3 ± i with accurate calculation using i² = -1. Choice D forgets the i in the solution: when the discriminant is negative, we have √(negative) = i√(positive), so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Conjugate pair shortcut: if one solution is a + bi, immediately write down a - bi as the other without recalculating! For real-coefficient quadratics, complex solutions ALWAYS come in conjugate pairs. If the quadratic formula gives you (6 + 2i)/2 = 3 + i from the + version, the - version automatically gives 3 - i. Same real part, flip the imaginary sign. Done!

Question 10

Solve $2x^2+3x+5=0$ . The discriminant is negative, so the solutions are complex. Use the quadratic formula and $i^2=-1$ , and express both solutions in $a+bi$ form.

$x=\dfrac{-3\pm i\sqrt{31}}{4}$ (correct answer)
$x=\dfrac{3\pm i\sqrt{31}}{4}$
$x=\dfrac{-3\pm i\sqrt{31}}{2}$
$x=\dfrac{-3\pm \sqrt{31}}{4}$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. The imaginary unit i is defined as i = √(-1), which means i² = -1: this allows us to take square roots of negative numbers. When we have √(-31), we can write it as i√31. Complex numbers have the form a + bi where a is the real part and b is the imaginary part. These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving 2x² + 3x + 5 = 0 using the quadratic formula: (1) Identify a = 2, b = 3, c = 5. (2) Calculate discriminant: b² - 4ac = (3)² - 4(2)(5) = 9 - 40 = -31. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-3 ± √(-31))/4 = (-3 ± i√31)/4. (5) The solutions are already in simplest form: x = (-3 + i√31)/4 and x = (-3 - i√31)/4. Choice A correctly expresses the solutions as x = (-3 ± i√31)/4 with accurate calculation using i² = -1. Choice D forgets the i in the solution: when the discriminant is negative, we have √(negative) = i√(positive), so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!

Question 11

Solve using the quadratic formula: $x^2 + 6x + 25 = 0$ . (Recall $i^2 = -1$ .) Express solutions in $a + bi$ form.

$x = -3 \pm 2i$
$x = -3 \pm 4$
$x = 3 \pm 4i$
$x = -3 \pm 4i$ (correct answer)

\sqrt{-1}

, which means

i^2 = -1

: this allows us to take square roots of negative numbers. When we have

\sqrt{-16}

, we can write it as

i \sqrt{16} = 4i

. Complex numbers have the form

a + bi

where a is the real part and b is the imaginary part (like

3 + 2i

-1 - 4i

). These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving

x^2 + 6x + 25 = 0

using the quadratic formula: (1) Identify a = 1, b = 6, c = 25. (2) Calculate discriminant:

b^2 - 4ac = 6^2 - 4(1)(25) = 36 - 100 = -64

. (3) Since discriminant is negative, solutions are complex. (4) Apply formula:

x = (-6 \pm \sqrt{-64})/2 = (-6 \pm 8i)/2

. (5) Simplify:

x = -6/2 \pm 8i/2 = -3 \pm 4i

. Solutions:

x = -3 + 4i

and

x = -3 - 4i

. Choice A correctly solves to get

x = -3 \pm 4i

with accurate calculation using

i^2 = -1

. Choice D forgets the i in the solution: when the discriminant is negative, we have

\sqrt{\text{negative}} = i \sqrt{\text{positive}}

, so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get

\sqrt{\text{negative number}}

, write it as

i \sqrt{\text{positive number}}

\sqrt{-16} = i \sqrt{16} = 4i

\sqrt{-5} = i \sqrt{5}

, (3) Simplify the entire expression to

a + bi

form by separating real parts from i-parts, (4) Write both solutions using

\pm

x = a \pm bi

means

x = a + bi

and

x = a - bi

. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative.

Question 12

Solve using the quadratic formula: $x^2 - 2x + 5 = 0$ . Express solutions in $a + bi$ form and use $i^2 = -1$ .

$x = 1 \pm 2$
$x = -1 \pm 2i$
$x = 1 \pm 2i$ (correct answer)
$x = 1 \pm i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. When a quadratic equation has discriminant b² - 4ac < 0 (negative), the quadratic formula gives √(negative), which means complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). We separate the real part (-b/(2a)) from the imaginary part (±√|b² - 4ac|/(2a))i to write in a + bi form. The solutions always come in conjugate pairs: a + bi and a - bi. Solving x² - 2x + 5 = 0 using the quadratic formula: (1) Identify a = 1, b = -2, c = 5. (2) Calculate discriminant: b² - 4ac = (-2)² - 4(1)(5) = 4 - 20 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (2 ± √(-16))/2 = (2 ± 4i)/2. (5) Simplify: x = 2/2 ± 4i/2 = 1 ± 2i. Solutions: x = 1 + 2i and x = 1 - 2i. Choice C correctly solves to get x = 1 ± 2i with accurate calculation using i² = -1. Choice D gives real solutions when the discriminant is negative. Check: b² - 4ac = 4 - 20 = -16. Negative discriminant means NO real solutions—only complex ones. When you get √(negative) in the formula, you can't ignore it or treat it as positive. You must include i! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -20 and you end up with no i in your answer, something went wrong.

Question 13

What are the solutions to $x^2-4x+5=0$ ? Write both solutions as complex numbers in $a\pm bi$ form (use $i^2=-1$ ).

$x=2\pm i$ (correct answer)
$x=-2\pm i$
$x=2\pm 2i$
$x=2\pm 1$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. Complex solutions appear in conjugate pairs for quadratics with real coefficients: if 2 + 3i is a solution, then 2 - 3i is automatically also a solution (same real part, opposite imaginary part). This pairing is guaranteed by the ± in the quadratic formula and has important implications: the sum is real (4 + 6i + 4 - 6i = 8), and the product is real ((2+3i)(2-3i) = 4 + 9 = 13), which is why quadratics with complex roots can still have real coefficients! Solving x² - 4x + 5 = 0 using the quadratic formula: (1) Identify a = 1, b = -4, c = 5. (2) Calculate discriminant: b² - 4ac = (-4)² - 4(1)(5) = 16 - 20 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (4 ± √(-4))/2 = (4 ± i√4)/2. (5) Simplify: x = 4/2 ± (√4/2)i = 2 ± 1i = 2 ± i. Solutions: x = 2 + i and x = 2 - i. Choice A correctly expresses in proper a + bi form x = 2 ± i with accurate calculation using i² = -1. Choice D forgets the i in the solution: when the discriminant is negative, we have √(negative) = i√(positive), so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Conjugate pair shortcut: if one solution is a + bi, immediately write down a - bi as the other without recalculating! For real-coefficient quadratics, complex solutions ALWAYS come in conjugate pairs. If the quadratic formula gives you (4 + 2i)/2 = 2 + i from the + version, the - version automatically gives 2 - i. Same real part, flip the imaginary sign. Done!

Question 14

Solve $x^2 + 2x + 2 = 0$ and write solutions in the form $a \pm bi$ . (Use $i^2 = -1$ .)

$x = 1 \pm i$
$x = -1 \pm i$ (correct answer)
$x = -1 \pm 2i$
$x = -1 \pm 1$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. The imaginary unit i is defined as i = √(-1), which means i² = -1: this allows us to take square roots of negative numbers. When we have √(-16), we can write it as i√16 = 4i. Complex numbers have the form a + bi where a is the real part and b is the imaginary part (like 3 + 2i or -1 - 4i). These aren't 'imaginary' in the sense of not existing—they're a real extension of the number system that makes every quadratic equation solvable! Solving x² + 2x + 2 = 0 using the quadratic formula: (1) Identify a = 1, b = 2, c = 2. (2) Calculate discriminant: b² - 4ac = (2)² - 4(1)(2) = 4 - 8 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-2 ± √(-4))/2 = (-2 ± 2i)/2. (5) Simplify: x = -2/2 ± 2i/2 = -1 ± i. Solutions: x = -1 + i and x = -1 - i. Choice B correctly solves to get x = -1 ± i with accurate calculation using i² = -1. Choice D forgets the i in the solution: when the discriminant is negative, we have √(negative) = i√(positive), so the i must appear in the answer. Without the i, these aren't complex numbers—they're just incorrect real numbers. The i is essential! Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get √(negative number), write it as i√(positive number): √(-16) = i√16 = 4i, √(-5) = i√5, (3) Simplify the entire expression to a + bi form by separating real parts from i-parts, (4) Write both solutions using ±: x = a ± bi means x = a + bi and x = a - bi. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative.

Question 15

Complete the square to solve $x^2+4x+13=0$ for complex solutions. Use $i^2=-1$ and write answers in $a\pm bi$ form.

$x=-2\pm 3i$ (correct answer)
$x=2\pm 3i$
$x=-2\pm 6i$
$x=-4\pm 3i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 4x + 13 = 0 by completing the square: (1) Move constant: x² + 4x = -13. (2) Add (b/2)²: x² + 4x + (4/2)² = -13 + 4 = -9. (3) Factor: (x + 2)² = -9. (4) Take square roots: x + 2 = ±√(-9) = ±i√9 = ±3i. (5) Solve: x = -2 ± 3i. The negative value after completing the square signals complex solutions! Choice A correctly expresses in proper a + bi form with accurate calculation using i² = -1. Choice B makes a sign error: the real part should be -2, not 2; with the ± and the i both appearing in solutions, it's easy to lose track of signs. Write out each solution separately: x = -2 + 3i AND x = -2 - 3i to make sure both signs are correct. Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get √(negative number), write it as i√(positive number): √(-16) = i√16 = 4i, √(-5) = i√5, (3) Simplify the entire expression to a + bi form by separating real parts from i-parts, (4) Write both solutions using ±: x = a ± bi means x = a + bi and x = a - bi. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative. Simplifying with i: remember that i² = -1 is the key rule. When you expand (a + bi)², you get a² + 2abi + (bi)² = a² + 2abi + b²i² = a² + 2abi - b² (since i² = -1). Combine reals and imaginaries separately: (a² - b²) + (2ab)i. This is how verification works—expand, use i² = -1, group real and imaginary parts.

Question 16

Determine the nature of the solutions to $x^2+4x+8=0$ , then solve. Because $b^2-4ac<0$ , write the solutions as complex numbers in $a\pm bi$ form using $i^2=-1$ .

$x=-2\pm 2i$ (correct answer)
$x=2\pm 2i$
$x=-2\pm i$
$x=-2\pm 2$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 4x + 8 = 0 using the quadratic formula: (1) Identify a = 1, b = 4, c = 8. (2) Calculate discriminant: b² - 4ac = (4)² - 4(1)(8) = 16 - 32 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-4 ± √(-16))/2 = (-4 ± i√16)/2. (5) Simplify: x = -4/2 ± (√16/2)i = -2 ± 2i. Solutions: x = -2 + 2i and x = -2 - 2i. Choice A correctly solves to get x = -2 ± 2i with accurate calculation using i² = -1. Choice C has the real and imaginary parts calculated wrong: from the quadratic formula x = (-b ± i√|discriminant|)/(2a), the real part is -b/(2a) = -4/2 = -2, not -2, and the imaginary part is √|discriminant|/(2a) = √16/2 = 4/2 = 2, not 1. Keep real and imaginary separate in your calculation! Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!

Question 17

What are the solutions to $x^2+4x+8=0$ ? Write the complex solutions in $a+bi$ form (use $i^2=-1$ ).

$x=-2\pm 2i$ (correct answer)
$x=2\pm 2i$
$x=-2\pm i$
$x=-2\pm 4i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 4x + 8 = 0 using the quadratic formula: (1) Identify a = 1, b = 4, c = 8. (2) Calculate discriminant: 4² - 4(1)(8) = 16 - 32 = -16. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-4 ± √(-16))/2 = (-4 ± i√16)/2. (5) Simplify: x = -4/2 ± (4i)/2 = -2 ± 2i. Solutions: x = -2 + 2i and x = -2 - 2i. Choice A correctly solves to get x = -2 ± 2i with accurate calculation using i² = -1. Choice B makes a sign error: the real part should be -2, not 2; with the ± and the i both appearing in solutions, it's easy to lose track of signs. Write out each solution separately: x = -2 + 2i AND x = -2 - 2i to make sure both signs are correct. Complex solution procedure: (1) Use quadratic formula or completing the square as usual, (2) When you get √(negative number), write it as i√(positive number): √(-16) = i√16 = 4i, √(-5) = i√5, (3) Simplify the entire expression to a + bi form by separating real parts from i-parts, (4) Write both solutions using ±: x = a ± bi means x = a + bi and x = a - bi. That's it! Complex solutions follow the same solving process, just with i appearing when discriminant is negative. The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -20 and you end up with no i in your answer, something went wrong.

Question 18

Solve using the quadratic formula: $x^2+2x+2=0$ . Because $b^2-4ac<0$ , write the complex solutions in $a\pm bi$ form using $i^2=-1$ .

$x=-1\pm 2i$
$x=1\pm i$
$x=-1\pm i$ (correct answer)
$x=-2\pm i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 2x + 2 = 0 using the quadratic formula: (1) Identify a = 1, b = 2, c = 2. (2) Calculate discriminant: b² - 4ac = (2)² - 4(1)(2) = 4 - 8 = -4. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-2 ± √(-4))/2 = (-2 ± i√4)/2. (5) Simplify: x = -2/2 ± (√4/2)i = -1 ± i. Solutions: x = -1 + i and x = -1 - i. Choice C correctly solves to get x = -1 ± i with accurate calculation using i² = -1. Choice A has the real and imaginary parts switched or calculated wrong: from the quadratic formula x = (-b ± i√|discriminant|)/(2a), the real part is -2/(2·1) = -1, not -1, and the imaginary part is √4/(2·1) = 1, not 2. Keep real and imaginary separate in your calculation! The discriminant is your early warning system: before solving, check b² - 4ac. Positive = two real solutions (no i needed). Zero = one real solution (perfect square). Negative = two complex solutions (i will appear). This tells you what to expect! If discriminant is -4 and you end up with no i in your answer, something went wrong.

Question 19

Complete the square to solve $x^2 + 4x + 13 = 0$ for complex solutions. Write answers in $a \pm bi$ form (use $i^2 = -1$ ).

$x = 2 \pm 3i$
$x = -2 \pm i$
$x = -2 \pm 3i$ (correct answer)
$x = -2 \pm 6i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 4x + 13 = 0 by completing the square: (1) Move constant: x² + 4x = -13. (2) Add (b/2)²: x² + 4x + 4 = -13 + 4 = -9. (3) Factor: (x + 2)² = -9. (4) Take square roots: x + 2 = ±√(-9) = ±3i. (5) Solve: x = -2 ± 3i. The negative value after completing the square signals complex solutions! Choice A correctly identifies the conjugate pair x = -2 ± 3i with accurate calculation using i² = -1. Choice D has the wrong imaginary coefficient: from (x + 2)² = -9, we get x + 2 = ±3i (since √9 = 3), so x = -2 ± 3i, not -2 ± 6i. The coefficient of i comes from √9 = 3, not 6. Always simplify under the radical first! Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!

Question 20

What are the solutions to $x^2 + 2x + 10 = 0$ ? Write as complex numbers in $a + bi$ form (use $i^2 = -1$ ).

$x = -1 \pm 3i$ (correct answer)
$x = 1 \pm 3i$
$x = -1 \pm i$
$x = -1 \pm 2i$

Explanation: This question tests your understanding of complex numbers and how to solve quadratic equations that have no real solutions but do have complex solutions involving the imaginary unit i. A negative discriminant tells you immediately: no real solutions exist, but two complex solutions do. Don't say 'no solution'—say 'no real solutions' or 'two complex solutions.' Complex solutions are valid mathematical solutions; they just don't correspond to x-intercepts on the standard real coordinate plane (the parabola doesn't cross the x-axis when discriminant is negative). Solving x² + 2x + 10 = 0 using the quadratic formula: (1) Identify a = 1, b = 2, c = 10. (2) Calculate discriminant: b² - 4ac = (2)² - 4(1)(10) = 4 - 40 = -36. (3) Since discriminant is negative, solutions are complex. (4) Apply formula: x = (-2 ± √(-36))/2 = (-2 ± 6i)/2. (5) Simplify: x = -2/2 ± 6i/2 = -1 ± 3i. Solutions: x = -1 + 3i and x = -1 - 3i. Choice A correctly solves to get x = -1 ± 3i with accurate calculation using i² = -1. Choice D has the wrong imaginary coefficient: from the quadratic formula x = (-2 ± 6i)/2, the imaginary part is 6i/2 = 3i, not 2i. Always simplify both the real and imaginary parts by dividing by 2a! Don't fear complex solutions: they're not 'worse' than real solutions, just different! In Algebra 1, you're learning that every quadratic has solutions—sometimes real (discriminant ≥ 0), sometimes complex (discriminant < 0). The complex solutions are just as valid mathematically. They don't represent x-intercepts on a standard graph, but they're still solutions to the equation!