Math II · N-CN.7

Solving Real-Coefficient Quadratic Equations with Complex Solutions

Complex solutions show students that a quadratic can have algebraic roots even when its graph never crosses the x-axis.

Concept Number and Quantity
Domain The Complex Number System
Read time 8 minutes

What this learning objective is really asking you to learn

This objective asks students to finish the quadratic story. Earlier, students learned that quadratic equations can be solved by factoring, square roots, completing the square, and the quadratic formula. They also learned that some quadratics have two real solutions, some have one real solution, and some appear to have no real solutions when viewed on the real number line. This objective says: if we allow complex numbers, the “no solution” case becomes “no real solution, but two complex solutions.”

A quadratic equation with real coefficients has the form

\[ax^2 + bx + c = 0\]

where \(a\), \(b\), and \(c\) are real numbers and \(a \ne 0\). The quadratic formula says

\[x = (-b ± \sqrt{b^2 - 4ac}) / (2a)\].

The expression under the square root, \(b^2 - 4ac\), is called the discriminant. It tells the type of solutions. If the discriminant is positive, there are two real solutions. If it is zero, there is one repeated real solution. If it is negative, there are no real solutions, but there are two complex solutions because square roots of negative numbers can be written using \(i\).

For example, \(x^2 + 4 = 0\) gives \(x^2 = -4\). In the real-number system, no number squared equals -4. But in the complex-number system, \(\sqrt{-4} = 2i\), so the solutions are \(x = 2i\) and \(x = -2i\).

A slightly richer example is \(x^2 - 6x + 13 = 0\). The discriminant is \(36 - 52 = -16\). The quadratic formula gives

\[x = (6 ± \sqrt{-16}) / 2\]

so

\[x = (6 ± 4i) / 2\]

which simplifies to

\[x = 3 ± 2i\].

Those two complex numbers are the roots. They are not x-intercepts, because x-intercepts must be real points on the coordinate plane where the graph crosses the real x-axis. Instead, they are algebraic solutions in the complex number system.

The objective is asking students to hold both ideas at once: graphically, a real quadratic with negative discriminant does not cross the x-axis; algebraically, it still has complex roots. That is a major mathematical maturity step. Students learn that “solution” depends on the number system being used.

Why students should learn this math

Students should learn this because it fixes a gap in their understanding of equations. If students stop at real numbers, they must say that some quadratics have no solutions. But algebra keeps pointing toward missing roots. Complex numbers let students complete the pattern. Every quadratic equation with real coefficients has two solutions when repeated roots and complex roots are counted properly.

This matters because mathematics is built by expanding systems when old systems are too small. Whole numbers could not solve \(3 + x = 1\), so negative numbers became necessary. Integers could not solve \(2x = 1\), so fractions became necessary. Rational numbers could not measure the diagonal of a unit square, so irrational numbers became necessary. Real numbers could not solve \(x^2 + 1 = 0\), so complex numbers became necessary. Students are watching the number system grow for a reason, not as a random complication.

There is also a practical modeling lesson. A quadratic graph tells a visual story. If it crosses the x-axis twice, the equation has two real zeros. If it touches once, it has one repeated real zero. If it stays above or below the x-axis, it has no real zeros. But the algebraic structure still has roots in the complex plane. This teaches students that one representation may hide information that another representation reveals.

In engineering, physics, and applied mathematics, complex roots are not decorative. They often describe oscillation, rotation, wave behavior, and stability. In differential equations and control systems, complex roots can indicate systems that vibrate, spiral, or oscillate while growing or decaying. Students do not need the full advanced theory in Math II, but they should know that complex solutions are not fake answers. They are the algebraic language of many real systems.

Even in ordinary school algebra, complex roots help students understand factoring. A quadratic like \(x^2 + 9\) cannot be factored into real linear factors, but over complex numbers it factors as \((x - 3i)(x + 3i)\). That reveals why complex numbers are part of polynomial structure. They let algebra continue working smoothly.

The deeper “why” is this: complex solutions teach students that mathematical systems have layers. The real graph is one layer. The algebraic equation is another. The number system determines what kinds of answers are visible. Learning complex quadratic solutions helps students stop thinking that “no x-intercept” means “nothing more can be said.” It means “no real zero,” not “no root at all.”

The historical machinery: why complex roots became unavoidable

Complex numbers entered mathematics partly because algebraists were trying to solve polynomial equations. Square roots of negative numbers appeared as strange intermediate objects, especially in work on cubic equations. At first, they seemed impossible. But mathematicians found that if they manipulated these quantities consistently, the results made sense and solved real problems.

Quadratics provide the simplest doorway. The equation \(x^2 + 1 = 0\) demands a number whose square is negative. The real-number system cannot supply one. Defining \(i\) by \(i^2 = -1\) creates a solution. Once that definition is accepted, more complicated negative square roots become manageable.

Over time, mathematicians realized that complex numbers were not merely symbolic tricks. They formed a coherent number system. They could be represented geometrically in the complex plane. They could be added, multiplied, and used to factor polynomials. This transformed algebra. Eventually, the Fundamental Theorem of Algebra clarified the power of the complex system: every nonconstant polynomial has a complex root. For quadratics, this means the roots are always there when complex numbers are allowed.

The history matters for students because it shows that complex roots were not invented to make school harder. They were invented, resisted, understood, and finally accepted because algebra needed them. The negative discriminant is not an annoyance. It is evidence that the real line is not large enough for all algebraic equations.

Where this fits in the big map of mathematics

This objective comes after students learn what \(i\) is and how to do arithmetic with complex numbers. It applies those tools to quadratic equations. The placement is important: students first need the number system, then the operations, then the solving application.

It connects backward to completing the square and the quadratic formula. Completing the square can also produce complex solutions. For example, \(x^2 - 6x + 13 = 0\) becomes \((x - 3)^2 + 4 = 0\), then \((x - 3)^2 = -4\), so \(x - 3 = ±2i\), giving \(x = 3 ± 2i\). This form connects directly to the vertex of the parabola. The graph has vertex \((3, 4)\), above the x-axis, so no real roots; algebra gives complex roots centered around real part 3.

It connects to graph interpretation. The discriminant tells whether a quadratic crosses, touches, or misses the x-axis. Complex roots appear in the “misses” case. Students should not expect to plot \(3 + 2i\) as a point on the ordinary x-axis. Instead, it belongs in the complex plane.

It connects forward to polynomial identities and the Fundamental Theorem of Algebra. Complex roots are not a special exception just for quadratics. They are part of the general root structure of polynomials.

It connects to advanced mathematics through oscillation and systems. Complex roots in later courses often indicate behavior that involves rotation or periodic motion. For now, students just need the seed idea: complex solutions carry information that real solutions cannot express.

The big-map role is completion. Quadratic solving becomes complete when students can handle the negative discriminant case.

How to execute the skill technically

The main method is the quadratic formula. Given \(ax^2 + bx + c = 0\), compute the discriminant \(D = b^2 - 4ac\). If \(D < 0\), write \(\sqrt{D}\) using \(i\).

Example: solve \(x^2 + 2x + 5 = 0\).

Here \(a = 1\), \(b = 2\), and \(c = 5\).

The discriminant is

\[D = 2^2 - 4(1)(5) = 4 - 20 = -16\].

Use the quadratic formula:

\[x = (-2 ± \sqrt{-16}) / 2\].

Since \(\sqrt{-16} = 4i\), we get

\[x = (-2 ± 4i) / 2\].

Divide both parts by 2:

\[x = -1 ± 2i\].

So the solutions are \(-1 + 2i\) and \(-1 - 2i\).

Another method is completing the square. Solve \(x^2 - 4x + 8 = 0\).

Move the constant:

\[x^2 - 4x = -8\].

Complete the square by adding 4 to both sides:

\[x^2 - 4x + 4 = -4\].

So

\[(x - 2)^2 = -4\].

Take square roots:

\[x - 2 = ±2i\].

Therefore

\[x = 2 ± 2i\].

Students should write answers in standard complex form \(a + bi\). The pair \(2 ± 2i\) means the two numbers \(2 + 2i\) and \(2 - 2i\).

A key pattern: for real-coefficient quadratics, nonreal complex roots come in conjugate pairs. If \(3 + 5i\) is a root, then \(3 - 5i\) is also a root. Students may not need to prove that yet, but they should notice it.

Another worked example: reading graph behavior and complex roots together

Consider the quadratic

\[f(x) = x^2 + 2x + 10\].

Complete the square:

\[f(x) = (x + 1)^2 + 9\].

This form says the parabola opens upward and has vertex \((-1, 9)\). Since the lowest output is 9, the graph never reaches zero. There are no real x-intercepts. Now solve algebraically:

\[x^2 + 2x + 10 = 0\].

Using the quadratic formula,

\[x = (-2 ± \sqrt{4 - 40}) / 2 = (-2 ± \sqrt{-36}) / 2\].

So

\[x = (-2 ± 6i) / 2 = -1 ± 3i\].

The real part of the roots, -1, matches the x-coordinate of the vertex. The imaginary part, 3i, is connected to the vertical distance from the vertex to the x-axis because the minimum value is 9 and \(\sqrt{9} = 3\). This is not something students must overgeneralize immediately, but it gives a valuable intuition: complex roots are not random decorations. They preserve information about a parabola that misses the real axis.

This is a strong way to teach the objective in an app. Show the parabola first. Let the student see it never crosses the x-axis. Then reveal the completed-square form and the complex roots. The student should learn to say: “No real intercepts, but two complex roots.”

How this changes the meaning of “solve”

Before complex numbers, solving a quadratic often meant finding where the graph crosses the x-axis. That is a good real-graph interpretation, but it is not the whole algebraic meaning. After complex numbers, solving means finding all numbers in the chosen number system that make the equation true. If the chosen system is real numbers, some quadratics have no solutions. If the chosen system is complex numbers, every quadratic has two solutions when multiplicity is counted.

That distinction is not pedantic. It is the beginning of careful mathematical language. A student should not say “there is no solution” when the accurate statement is “there is no real solution.” This precision prepares students for later work where domain, range, number system, and assumptions determine what answers are allowed.

Common misconceptions and how to avoid them

One common mistake is saying a quadratic with a negative discriminant has “no solution.” More precisely, it has no real solution. In the complex number system, it has complex solutions.

Another mistake is simplifying \(\sqrt{-16}\) as -4. That is false. Since \((-4)^2 = 16\), not -16, the correct simplification is 4i.

A third mistake is forgetting to divide both the real and imaginary parts by 2a in the quadratic formula. For example, \((-2 ± 4i)/2\) becomes \(-1 ± 2i\), not \(-2 ± 2i\).

A fourth mistake is expecting complex roots to appear as x-intercepts on a real graph. They do not. They are solutions in the complex plane, not real x-axis crossing points.

A fifth mistake is treating \(±\) as one answer. It represents two solutions: one with plus and one with minus.

The big takeaway

This objective completes the quadratic-solving picture. A negative discriminant means no real roots, but complex numbers allow the equation to have solutions. Students learn to use \(i\), simplify negative square roots, and express answers in \(a + bi\) form. The larger lesson is that algebra sometimes points beyond the number system students already know, and expanding the number system reveals structure that was hidden before.

Problem Library

Problems in the App From This Objective

168 problems across 12 archetypes in the app.

evaluate negative discriminant.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Use the discriminant to identify whether quadratic x^2+4x+8=0 has no real solutions.

Open in simulator
Problem 2

Use the discriminant to identify whether quadratic 2x^2-3x+5=0 has no real solutions.

Problem 3

Use the discriminant to identify whether quadratic x^2-6x+9=0 has no real solutions.

Problem 4

Use the discriminant to identify whether quadratic x^2-5x+6=0 has no real solutions.

Problem 5

Use the discriminant to identify whether quadratic 3x^2+2x+1=0 has no real solutions.

Problem 6

Use the discriminant to identify whether quadratic x^2+x+1=0 has no real solutions.

Problem 7

Use the discriminant to identify whether quadratic 5x^2-4x+2=0 has no real solutions.

Problem 8

Use the discriminant to identify whether quadratic x^2+2x+1=0 has no real solutions.

Problem 9

Use the discriminant to identify whether quadratic 4x^2-4x+1=0 has no real solutions.

Problem 10

Use the discriminant to identify whether quadratic 9x^2+6x+1=0 has no real solutions.

Problem 11

Use the discriminant to identify whether quadratic x^2-7x+10=0 has no real solutions.

Problem 12

Use the discriminant to identify whether quadratic 2x^2+5x-3=0 has no real solutions.

Problem 13

Use the discriminant to identify whether quadratic x^2+3x-4=0 has no real solutions.

Problem 14

Use the discriminant to identify whether quadratic 3x^2+10x+3=0 has no real solutions.

Problem 15

Use the discriminant to identify whether quadratic x^2-2x+3=0 has no real solutions.

simplify `sqrt(-a)`.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Solve quadratic x^2+25=0 with pure imaginary solutions by square roots.

Problem 17

Solve quadratic x^2+12=0 with pure imaginary solutions by square roots.

Problem 18

Solve quadratic 3x^2+27=0 with pure imaginary solutions by square roots.

Problem 19

Solve quadratic x^2=-7 with pure imaginary solutions by square roots.

Problem 20

Solve quadratic x^2+4=0 with pure imaginary solutions by square roots.

Problem 21

Solve quadratic x^2+11=0 with pure imaginary solutions by square roots.

Problem 22

Solve quadratic x^2+20=0 with pure imaginary solutions by square roots.

Open in simulator
Problem 23

Solve quadratic 2x^2+50=0 with pure imaginary solutions by square roots.

Problem 24

Solve quadratic 4x^2+36=0 with pure imaginary solutions by square roots.

Problem 25

Solve quadratic 5x^2+125=0 with pure imaginary solutions by square roots.

Problem 26

Solve quadratic x^2=-100 with pure imaginary solutions by square roots.

Problem 27

Solve quadratic x^2=-13 with pure imaginary solutions by square roots.

Problem 28

Solve quadratic x^2=-48 with pure imaginary solutions by square roots.

Problem 29

Solve quadratic 2x^2+18=0 with pure imaginary solutions by square roots.

Problem 30

Solve quadratic x^2+75=0 with pure imaginary solutions by square roots.

apply quadratic formula and simplify imaginary radical.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Solve quadratic x^2-4x+8=0 using the quadratic formula.

Problem 32

Solve quadratic x^2+2x+5=0 using the quadratic formula.

Problem 33

Solve quadratic 2x^2-4x+10=0 using the quadratic formula.

Problem 34

Solve quadratic x^2+6x+13=0 using the quadratic formula.

Problem 35

Solve quadratic x^2+1=0 using the quadratic formula.

Problem 36

Solve quadratic x^2+2x+2=0 using the quadratic formula.

Problem 37

Solve quadratic x^2+4x+5=0 using the quadratic formula.

Open in simulator
Problem 38

Solve quadratic x^2-6x+10=0 using the quadratic formula.

Problem 39

Solve quadratic x^2+9=0 using the quadratic formula.

Problem 40

Solve quadratic x^2+2x+10=0 using the quadratic formula.

Problem 41

Solve quadratic x^2+4x+13=0 using the quadratic formula.

Problem 42

Solve quadratic x^2-2x+10=0 using the quadratic formula.

Problem 43

Solve quadratic x^2+16=0 using the quadratic formula.

Problem 44

Solve quadratic x^2+2x+17=0 using the quadratic formula.

Problem 45

Solve quadratic x^2-4x+20=0 using the quadratic formula.

isolate squared binomial equal to negative value.
12 problems Warmup Practice Mixed Review Assessment
Problem 46

Complete the square to solve (x-3)^2=-16.

Problem 47

Complete the square to solve x^2+6x+13=0.

Problem 48

Complete the square to solve x^2-10x+29=0.

Problem 49

Complete the square to solve (x+1)^2=-7.

Problem 50

Complete the square to solve (x-5)^2=-9.

Problem 51

Complete the square to solve (x+2)^2=-5.

Problem 52

Complete the square to solve x^2-8x+41=0.

Problem 53

Complete the square to solve x^2+2x+4=0.

Problem 54

Complete the square to solve (x+4)^2=-36.

Problem 55

Complete the square to solve x^2-4x+5=0.

Problem 56

Complete the square to solve (x-6)^2=-11.

Open in simulator
Problem 57

Complete the square to solve x^2+10x+27=0.

write `a +/- bi`.
12 problems Warmup Practice Mixed Review Assessment
Problem 58

Express complex solutions x=(4 +/- sqrt(-36))/2 in standard form.

Problem 59

Express complex solutions x=(-6 +/- sqrt(-16))/2 in standard form.

Problem 60

Express complex solutions x=(2 +/- sqrt(-20))/2 in standard form.

Problem 61

Express complex solutions x=(-8 +/- sqrt(-12))/4 in standard form.

Problem 62

Express complex solutions x=(10 +/- sqrt(-100))/2 in standard form.

Problem 63

Express complex solutions x=(12 +/- sqrt(-28))/2 in standard form.

Open in simulator
Problem 64

Express complex solutions x=(5 +/- sqrt(-16))/2 in standard form.

Problem 65

Express complex solutions x=(7 +/- sqrt(-20))/2 in standard form.

Problem 66

Express complex solutions x=(-10 +/- sqrt(-64))/2 in standard form.

Problem 67

Express complex solutions x=(-18 +/- sqrt(-72))/6 in standard form.

Problem 68

Express complex solutions x=(15 +/- sqrt(-75))/5 in standard form.

Problem 69

Express complex solutions x=(9 +/- sqrt(-48))/6 in standard form.

substitute and simplify complex arithmetic.
15 problems Warmup Practice Mixed Review Assessment
Problem 70

Verify complex solution 2+3i in quadratic x^2-4x+13=0.

Problem 71

Verify complex solution -1+2i in quadratic x^2+2x+5=0.

Problem 72

Verify complex solution 4i in quadratic x^2+16=0.

Problem 73

Verify complex solution 3-i in quadratic x^2-6x+10=0.

Problem 74

Verify complex solution 1+i in quadratic x^2-2x+2=0.

Open in simulator
Problem 75

Verify complex solution 1-i in quadratic x^2-2x+2=0.

Problem 76

Verify complex solution -2+i in quadratic x^2+4x+5=0.

Problem 77

Verify complex solution -2-i in quadratic x^2+4x+5=0.

Problem 78

Verify complex solution 5i in quadratic x^2+25=0.

Problem 79

Verify complex solution -5i in quadratic x^2+25=0.

Problem 80

Verify complex solution 3+2i in quadratic x^2-6x+13=0.

Problem 81

Verify complex solution 3-2i in quadratic x^2-6x+13=0.

Problem 82

Verify complex solution -1+3i in quadratic x^2+2x+10=0.

Problem 83

Verify complex solution -1-3i in quadratic x^2+2x+10=0.

Problem 84

Verify complex solution 2-5i in quadratic x^2-4x+29=0.

interpret negative discriminant graphically.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Connect complex solutions to graph feature parabola opens upward and stays above the x-axis.

Open in simulator
Problem 86

Connect complex solutions to graph feature quadratic has discriminant -20.

Problem 87

Connect complex solutions to graph feature parabola opens downward and stays below the x-axis.

Problem 88

Connect complex solutions to graph feature vertex is on the x-axis.

Problem 89

Connect complex solutions to graph feature the graph does not intersect the x-axis.

Problem 90

Connect complex solutions to graph feature parabola has a minimum value of 3.

Problem 91

Connect complex solutions to graph feature parabola has a maximum value of -5.

Problem 92

Connect complex solutions to graph feature the quadratic function never equals zero for any real x.

Problem 93

Connect complex solutions to graph feature the discriminant of the quadratic equation is -16.

Problem 94

Connect complex solutions to graph feature the parabola crosses the x-axis at two distinct points.

Problem 95

Connect complex solutions to graph feature the graph touches the x-axis at exactly one point.

Problem 96

Connect complex solutions to graph feature the range of the quadratic function is `[2, infinity)`.

build conjugate factors and multiply.
15 problems Warmup Practice Mixed Review Assessment
Problem 97

Find a quadratic equation with complex conjugate roots 2+3i and 2-3i.

Problem 98

Find a quadratic equation with complex conjugate roots -1+4i and -1-4i.

Problem 99

Find a quadratic equation with complex conjugate roots 5+i and 5-i.

Problem 100

Find a quadratic equation with complex conjugate roots 1/2+i and 1/2-i.

Problem 101

Find a quadratic equation with complex conjugate roots 3+2i and 3-2i.

Problem 102

Find a quadratic equation with complex conjugate roots -2+5i and -2-5i.

Problem 103

Find a quadratic equation with complex conjugate roots 4i and -4i.

Problem 104

Find a quadratic equation with complex conjugate roots 1+i and 1-i.

Problem 105

Find a quadratic equation with complex conjugate roots 4+3i and 4-3i.

Problem 106

Find a quadratic equation with complex conjugate roots -3+i and -3-i.

Problem 107

Find a quadratic equation with complex conjugate roots 1/3+2/3i and 1/3-2/3i.

Problem 108

Find a quadratic equation with complex conjugate roots i and -i.

Open in simulator
Problem 109

Find a quadratic equation with complex conjugate roots 5+2i and 5-2i.

Problem 110

Find a quadratic equation with complex conjugate roots -4+2i and -4-2i.

Problem 111

Find a quadratic equation with complex conjugate roots 2+1/2i and 2-1/2i.

distinguish two real, one real, two complex.
12 problems Warmup Practice Mixed Review Assessment
Problem 112

Classify solutions from discriminant value 25.

Problem 113

Classify solutions from discriminant value 0.

Problem 114

Classify solutions from discriminant value -9.

Problem 115

Classify solutions from discriminant value 3.

Problem 116

Classify solutions from discriminant value 16.

Problem 117

Classify solutions from discriminant value 7.

Open in simulator
Problem 118

Classify solutions from discriminant value 100.

Problem 119

Classify solutions from discriminant value 13.

Problem 120

Classify solutions from discriminant value -1.

Problem 121

Classify solutions from discriminant value -25.

Problem 122

Classify solutions from discriminant value 1.

Problem 123

Classify solutions from discriminant value 2.

connect formula structure to +/- imaginary part.
15 problems Warmup Practice Mixed Review Assessment
Problem 124

Explain why real-coefficient quadratic x^2-4x+13=0 has complex roots in conjugate pairs.

Open in simulator
Problem 125

Explain why real-coefficient quadratic x^2+2x+5=0 has complex roots in conjugate pairs.

Problem 126

Explain why real-coefficient quadratic ax^2+bx+c=0 with real a,b,c and negative discriminant has complex roots in conjugate pairs.

Problem 127

Explain why real-coefficient quadratic x^2+1=0 has complex roots in conjugate pairs.

Problem 128

Explain why real-coefficient quadratic x^2-2x+2=0 has complex roots in conjugate pairs.

Problem 129

Explain why real-coefficient quadratic x^2-6x+10=0 has complex roots in conjugate pairs.

Problem 130

Explain why real-coefficient quadratic x^2+4x+5=0 has complex roots in conjugate pairs.

Problem 131

Explain why real-coefficient quadratic x^2-2x+5=0 has complex roots in conjugate pairs.

Problem 132

Explain why real-coefficient quadratic x^2+4=0 has complex roots in conjugate pairs.

Problem 133

Explain why real-coefficient quadratic x^2+9=0 has complex roots in conjugate pairs.

Problem 134

Explain why real-coefficient quadratic 2x^2-2x+1=0 has complex roots in conjugate pairs.

Problem 135

Explain why real-coefficient quadratic x^2+2x+10=0 has complex roots in conjugate pairs.

Problem 136

Explain why real-coefficient quadratic 4x^2-16x+17=0 has complex roots in conjugate pairs.

Problem 137

Explain why real-coefficient quadratic x^2-2x+17=0 has complex roots in conjugate pairs.

Problem 138

Explain why real-coefficient quadratic 9x^2-6x+10=0 has complex roots in conjugate pairs.

select square roots, completing square, or formula.
15 problems Warmup Practice Mixed Review Assessment
Problem 139

Choose the best method to solve quadratic x^2+25=0.

Problem 140

Choose the best method to solve quadratic x^2+6x+13=0.

Problem 141

Choose the best method to solve quadratic 3x^2-2x+7=0.

Problem 142

Choose the best method to solve quadratic (x-4)^2+9=0.

Problem 143

Choose the best method to solve quadratic 2x^2 - 18 = 0.

Problem 144

Choose the best method to solve quadratic (x+5)^2 - 16 = 0.

Problem 145

Choose the best method to solve quadratic 5x^2 = 45.

Problem 146

Choose the best method to solve quadratic 3(x-1)^2 = 27.

Problem 147

Choose the best method to solve quadratic x^2 - 8x + 20 = 0.

Open in simulator
Problem 148

Choose the best method to solve quadratic x^2 + 10x - 3 = 0.

Problem 149

Choose the best method to solve quadratic x^2 - 2x + 5 = 0.

Problem 150

Choose the best method to solve quadratic 2x^2 + 5x - 3 = 0.

Problem 151

Choose the best method to solve quadratic x^2 + 3x + 1 = 0.

Problem 152

Choose the best method to solve quadratic 4x^2 - 5x + 1 = 0.

Problem 153

Choose the best method to solve quadratic 5x^2 + 7x + 2 = 0.

catch discriminant, square-root, sign, and standard-form mistakes.
15 problems Warmup Practice Mixed Review Assessment
Problem 154

Correct the complex quadratic solution error in x^2+9=0 gives x=3.

Problem 155

Correct the complex quadratic solution error in For x^2-4x+8=0, discriminant is 16-32=-16, so x=2+/-sqrt(16)/2.

Problem 156

Correct the complex quadratic solution error in x=(-6 +/- sqrt(-16))/2 is simplified to -3+/-4i.

Problem 157

Correct the complex quadratic solution error in A negative discriminant means the quadratic has no solutions.

Problem 158

Correct the complex quadratic solution error in x^2+16=0 gives x=4.

Problem 159

Correct the complex quadratic solution error in 2x^2+18=0 gives x=3.

Open in simulator
Problem 160

Correct the complex quadratic solution error in For x^2+2x+5=0, discriminant is 4-20=16, so x=(-2+/-4)/2.

Problem 161

Correct the complex quadratic solution error in For 3x^2-6x+9=0, discriminant is 36-108=72, so x=(6+/-sqrt(72))/6.

Problem 162

Correct the complex quadratic solution error in x=(-8 +/- sqrt(-36))/2 is simplified to -4+/-sqrt(-36).

Problem 163

Correct the complex quadratic solution error in x=(4 +/- sqrt(-20))/2 is simplified to 2+/-sqrt(-5).

Problem 164

Correct the complex quadratic solution error in If the discriminant is negative, the quadratic has one real solution.

Problem 165

Correct the complex quadratic solution error in x^2=-25 gives x=+/-5.

Problem 166

Correct the complex quadratic solution error in x^2-6x+13=0 leads to x=(6 +/- sqrt(36-52))/2 = (6 +/- sqrt(16))/2 = 3+/-2.

Problem 167

Correct the complex quadratic solution error in x=(-10 +/- sqrt(-100))/5 is simplified to -2+/-2.

Problem 168

Correct the complex quadratic solution error in The solutions to x^2+2x+2=0 are -1+i and -1-i. These are real solutions.