Math II · A-REI.4.b

Solving Quadratic Equations by the Best Available Method, Including Complex Solutions

This objective teaches students how to handle one of the most common nonlinear equation types in the world. Quadratics model area, motion, profit, design, and many “rise then fall” or “minimum/maximum” situations.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 10 minutes

What this learning objective is really asking you to learn

This objective asks students to become flexible quadratic solvers. A quadratic equation is an equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\) is not zero. The squared term makes the equation fundamentally different from a linear equation. A linear equation usually has one solution. A quadratic equation can have two real solutions, one repeated real solution, or no real solutions. Once complex numbers are allowed, every quadratic has two solutions when multiplicity is counted.

The standard names several methods: inspection, square roots, completing the square, the quadratic formula, and factoring. That list is not random. Each method matches a particular structure. A student who sees \(x^2 = 49\) should not need the quadratic formula. Inspection or square roots solve it immediately. A student who sees \((x - 3)^2 = 20\) should take square roots. A student who sees \((x - 5)(x + 2) = 0\) should use the zero product property. A student who sees \(x^2 + 6x + 9 = 0\) may recognize a perfect square. A student who sees \(2x^2 - 7x + 3 = 0\) may factor. A student who sees \(5x^2 + 2x + 9 = 0\) may need the quadratic formula and may discover complex solutions.

Solving quadratics is also connected to graphs. If a quadratic equation is written as \(f(x) = 0\), the real solutions are the x-intercepts of the graph \(y = f(x)\). If the graph crosses the x-axis twice, there are two real solutions. If it touches the x-axis at the vertex, there is one repeated real solution. If it never reaches the x-axis, there are no real solutions. Complex solutions exist algebraically, but they do not appear as x-intercepts on the real coordinate plane.

The quadratic formula is the universal method. For \(ax^2 + bx + c = 0\), the solutions are \(x = (-b ± \sqrt{b^2 - 4ac})/(2a)\). The expression under the square root, \(b^2 - 4ac\), is called the discriminant. It discriminates among solution types. If it 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.

Complex solutions are written using \(i\), where \(i^2 = -1\). If the quadratic formula gives \(\sqrt{-16}\), students rewrite it as 4i. If it gives \(\sqrt{-20}\), students rewrite it as \(2\sqrt{5}i\). A solution such as \(x = 3 ± 2i\) means there are two solutions: \(3 + 2i\) and \(3 - 2i\). These are complex conjugates. For quadratics with real coefficients, non-real complex solutions always come in conjugate pairs.

This objective is not just about procedure. It is about judgment. Students must choose a method appropriate to the initial form. They must preserve equivalence. They must check whether solutions make sense in context. They must understand that an equation may have algebraic solutions that are not meaningful in a real-world model. For example, a projectile equation might have two real time solutions for a certain height: one on the way up and one on the way down. But if a solution is negative time, it may be outside the modeled situation. If an area problem gives a negative length, that solution must be rejected in context.

A strong student can explain the same quadratic in multiple representations. In factored form, the zeros are visible. In vertex form, the maximum or minimum is visible. In standard form, the y-intercept may be visible and the quadratic formula is available. Different forms reveal different information. Solving is partly the art of moving to the form that makes the desired information easiest to see.

Why students should learn this math

Students should learn quadratic solving because quadratics are the first major nonlinear family students can solve in a systematic way. Linear equations model constant change, but the world is not only linear. Quadratics appear when quantities are multiplied, when areas are calculated, when objects move under constant acceleration, when profit depends on price and demand, when distances are squared, and when designs involve curved shapes.

Projectile motion is a classic example. The height of an object thrown upward can often be approximated by a quadratic function of time. Solving a quadratic can answer when the object reaches the ground, when it reaches a certain height, or whether it clears an obstacle. The two-solution possibility has real meaning: the object may pass a certain height once on the way up and again on the way down.

Area problems also create quadratics. If a rectangle's length and width both depend on the same unknown, the area equation may become quadratic. Solving tells which dimensions produce a desired area. Revenue and profit models can also be quadratic when price affects quantity sold. A business may need to find break-even points or the price range where profit is positive.

Quadratic solving also teaches students that mathematical tools should be chosen intelligently. Many students want one procedure for everything because it feels safe. But real mathematical competence means recognizing structure. Factoring is fast when the equation factors. Square roots are fast when the equation is already a square equal to a number. Completing the square reveals vertex structure. The quadratic formula handles all cases but can be slower and more error-prone if a simpler method exists. Method selection is a form of mathematical maturity.

Complex solutions matter because they show that the real number system is not the end of algebra. Students first meet negative numbers because subtraction demands them. They meet fractions because division demands them. They meet irrational numbers because square roots demand them. They meet complex numbers because equations like \(x^2 + 1 = 0\) demand them. The number system expands when existing equations ask questions the old system cannot answer.

Even when complex solutions do not represent direct physical measurements in a beginning context, they are not fake. They become essential in advanced algebra, electrical engineering, signal processing, quantum mechanics, control systems, fluid dynamics, and many areas of mathematics. In Math II, students only need the doorway: recognize when the quadratic formula produces a negative under the square root and write the solutions in \(a ± bi\) form.

There is also a confidence reason. The quadratic formula is often treated as a monster to memorize. Students who understand the methods behind quadratic solving can approach it calmly. They know that the formula is a tool, not a ritual. They know how to check the discriminant before plunging into computation. They know how to connect the answer to a graph. That reduces fear and builds durable understanding.

The historical machinery behind this idea

Quadratic equations are ancient. Problems involving squares, rectangles, and unknown lengths appear in Babylonian mathematics thousands of years ago. These problems were often solved using methods equivalent to completing the square. The notation was not modern, but the reasoning was recognizable: arrange areas, complete missing pieces, and find the unknown length.

Greek geometry also contributed to quadratic thinking through area relationships and geometric constructions. Later, mathematicians in the Islamic world, including al-Khwarizmi, organized quadratic equation solving into systematic cases. Because negative numbers and symbolic notation were not used the same way as today, different forms of quadratic equations were treated separately. For example, \(x^2 + bx = c\) and \(x^2 = bx + c\) might be considered different cases. Modern notation lets us combine them into \(ax^2 + bx + c = 0\).

The quadratic formula became possible in its modern compact form only after symbolic algebra matured. Once letters could stand for general coefficients, mathematicians could solve the general quadratic instead of solving one numerical example at a time. Completing the square on \(ax^2 + bx + c = 0\) produced the formula students use today.

Complex numbers entered the story because polynomial equations pushed mathematicians beyond real numbers. At first, square roots of negative numbers seemed suspicious or impossible. But as algebra developed, especially through work on cubic equations in the Renaissance, mathematicians found that these “imaginary” quantities could appear in intermediate steps and still lead to real answers. Rafael Bombelli is often associated with early systematic rules for complex numbers. Over time, complex numbers became accepted as legitimate mathematical objects with deep geometric and practical meaning.

This history can help students see that solving quadratics is not a school invention. It is part of humanity's long attempt to solve area, motion, and relationship problems. The methods students learn are streamlined versions of ideas developed over centuries.

Technical execution: choosing the right method

A good quadratic solver begins by putting the equation into a useful form. If the equation is not already set equal to zero, decide whether setting it to zero will help. Standard form \(ax^2 + bx + c = 0\) is useful for factoring and the quadratic formula. Squared form \((x - h)^2 = k\) is useful for square roots. Factored form \(a(x - r)(x - s) = 0\) is useful for reading zeros.

Inspection works when the solution is immediately visible. For \(x^2 = 49\), students can see \(x = 7\) and \(x = -7\). For \(x^2 = 0\), the only solution is \(x = 0\). Inspection should not mean guessing wildly; it means recognizing a simple structure.

Taking square roots works when the equation can be written as a squared expression equal to a number. For \((x + 4)^2 = 25\), take square roots: \(x + 4 = ±5\), giving \(x = 1\) or \(x = -9\). Students must include both signs unless context eliminates one. For \((2x - 1)^2 = 18\), take square roots: \(2x - 1 = ±\sqrt{18}\), so \(x = (1 ± \sqrt{18})/2\), which simplifies to \(x = (1 ± 3\sqrt{2})/2\).

Factoring works when the quadratic can be written as a product. For \(x^2 - 5x + 6 = 0\), factor to \((x - 2)(x - 3) = 0\), so \(x = 2\) or \(x = 3\). The zero product property is the key: if a product equals zero, at least one factor must equal zero. Factoring is also useful for interpreting graphs because factored form shows x-intercepts.

Completing the square works for any quadratic but is especially useful when vertex form is needed. For \(x^2 + 4x - 1 = 0\), move the constant: \(x^2 + 4x = 1\). Add 4: \((x + 2)^2 = 5\). Then \(x = -2 ± \sqrt{5}\). This method is also the foundation of the quadratic formula.

The quadratic formula works for every quadratic in standard form. For \(2x^2 - 3x - 5 = 0\), use \(a = 2\), \(b = -3\), \(c = -5\). The discriminant is \((-3)^2 - 4(2)(-5) = 9 + 40 = 49\). Then \(x = (3 ± 7)/4\), giving \(x = 10/4 = 5/2\) or \(x = -4/4 = -1\). The formula gives the same solutions factoring would give, but it works even when factoring is not easy.

For complex solutions, consider \(x^2 + 4x + 13 = 0\). Here \(a = 1\), \(b = 4\), \(c = 13\). The discriminant is \(16 - 52 = -36\). The formula gives \(x = (-4 ± \sqrt{-36})/2 = (-4 ± 6i)/2 = -2 ± 3i\). There are no real x-intercepts, but there are two complex solutions.

Students should use the discriminant strategically. Before completing the whole formula, compute \(b^2 - 4ac\). If it is positive, expect two real solutions. If zero, expect one repeated real solution. If negative, expect complex solutions. This expectation helps catch mistakes.

Interpreting solutions in context

Quadratic solutions are values that make the equation true. But in applications, they must be interpreted. Suppose a height equation gives two positive times when the object is 20 feet high. Both may be meaningful: one time on the way up and one on the way down. Suppose a revenue equation gives two prices that break even. Those may represent lower and upper break-even prices, with profit between them. Suppose an area equation gives one positive and one negative dimension. The negative dimension is rejected because it does not make sense physically.

Students should also connect algebraic solutions to graphs. If a quadratic has two real roots, the parabola crosses the x-axis twice. If it has one repeated root, the vertex touches the x-axis. If it has complex roots, the parabola does not cross the x-axis. This visual connection makes the discriminant meaningful instead of mysterious.

Where this fits in the big map of mathematics

This objective is central to Integrated Math II. It depends on polynomial arithmetic from Objective 060, equation creation from Objective 061, graphing relationships from Objective 062, formula rearrangement from Objective 063, and completing the square from Objective 064. It prepares students for linear-quadratic systems, quadratic functions, complex number arithmetic, conic sections, and polynomial roots.

In the larger map, quadratics are the first major example of a polynomial family. Later, students will study higher-degree polynomials, rational functions, and more advanced equations. The habits developed here—choose a form, select a method, interpret roots, check the number system—carry forward.

Common student traps and how to avoid them

One trap is forgetting that square roots produce two solutions. \(x^2 = 49\) has \(x = 7\) and \(x = -7\). The principal square root symbol \(\sqrt{49}\) means 7, but solving an equation with a square requires both possibilities.

A second trap is using the quadratic formula with wrong signs. Students should write \(a\), \(b\), and \(c\) clearly before substituting. Parentheses are essential, especially when \(b\) is negative.

A third trap is assuming no real solution means no solution at all. It means no real solution. Complex solutions may still exist and can be written using \(i\).

A fourth trap is using factoring when the quadratic does not factor nicely and then forcing incorrect factors. Factoring is powerful, but it should not become guessing under pressure. The quadratic formula is available when factoring is not clean.

A fifth trap is ignoring context. Algebra may give two answers, but the situation may accept one, both, or neither. Always return to the original question.

Problem Library

Problems in the App From This Objective

213 problems across 15 archetypes in the app.

factor and use zero-product property.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Solve x^2+7x+12=0 by factoring the monic trinomial.

Problem 2

Solve x^2-x-20=0 by factoring the monic trinomial.

Problem 3

Solve x^2-9x+20=0 by factoring the monic trinomial.

Problem 4

Solve x^2+2x-15=0 by factoring the monic trinomial.

Problem 5

Solve x^2-3x+2=0 by factoring the monic trinomial.

Open in simulator
Problem 6

Solve x^2-x-2=0 by factoring the monic trinomial.

Problem 7

Solve x^2+3x+2=0 by factoring the monic trinomial.

Problem 8

Solve x^2-13x+42=0 by factoring the monic trinomial.

Problem 9

Solve x^2-x-42=0 by factoring the monic trinomial.

Problem 10

Solve x^2+13x+42=0 by factoring the monic trinomial.

Problem 11

Solve x^2-5x=0 by factoring the monic trinomial.

Problem 12

Solve x^2+5x=0 by factoring the monic trinomial.

Problem 13

Solve x^2-9=0 by factoring the monic trinomial.

Problem 14

Solve x^2-17x+72=0 by factoring the monic trinomial.

Problem 15

Solve x^2-6x-16=0 by factoring the monic trinomial.

factor with leading coefficient not 1.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Solve 2x^2+7x+3=0 by factoring a non-monic trinomial.

Open in simulator
Problem 17

Solve 3x^2-10x+8=0 by factoring a non-monic trinomial.

Problem 18

Solve 4x^2-12x+9=0 by factoring a non-monic trinomial.

Problem 19

Solve 6x^2+x-2=0 by factoring a non-monic trinomial.

Problem 20

Solve 5x^2+11x+2=0 by factoring a non-monic trinomial.

Problem 21

Solve 6x^2-17x+5=0 by factoring a non-monic trinomial.

Problem 22

Solve 4x^2-x-3=0 by factoring a non-monic trinomial.

Problem 23

Solve 6x^2+5x-6=0 by factoring a non-monic trinomial.

Problem 24

Solve 7x^2+30x+8=0 by factoring a non-monic trinomial.

Problem 25

Solve 10x^2-11x+3=0 by factoring a non-monic trinomial.

Problem 26

Solve 9x^2-9x-4=0 by factoring a non-monic trinomial.

Problem 27

Solve 8x^2+18x-5=0 by factoring a non-monic trinomial.

Problem 28

Solve 4x^2+3x-10=0 by factoring a non-monic trinomial.

Problem 29

Solve 6x^2-17x-3=0 by factoring a non-monic trinomial.

Problem 30

Solve 25x^2-20x+4=0 by factoring a non-monic trinomial.

recognize conjugate factors.
12 problems Warmup Practice Mixed Review Assessment
Problem 31

Solve the difference-of-squares equation x^2-49=0.

Problem 32

Solve the difference-of-squares equation 4x^2-81=0.

Problem 33

Solve the difference-of-squares equation (x-3)^2-25=0.

Problem 34

Solve the difference-of-squares equation 9x^2-16=0.

Open in simulator
Problem 35

Solve the difference-of-squares equation x^2-100=0.

Problem 36

Solve the difference-of-squares equation 25x^2-1=0.

Problem 37

Solve the difference-of-squares equation (x-1)^2-81=0.

Problem 38

Solve the difference-of-squares equation 36x^2-49=0.

Problem 39

Solve the difference-of-squares equation x^2-121=0.

Problem 40

Solve the difference-of-squares equation (x+2)^2-16=0.

Problem 41

Solve the difference-of-squares equation 49x^2-4=0.

Problem 42

Solve the difference-of-squares equation (3x-5)^2-64=0.

isolate squared expression and include plus/minus.
12 problems Warmup Practice Mixed Review Assessment
Problem 43

Solve x^2=36 by taking square roots.

Open in simulator
Problem 44

Solve (x-4)^2=9 by taking square roots.

Problem 45

Solve 2(x+1)^2=18 by taking square roots.

Problem 46

Solve (x+5)^2=7 by taking square roots.

Problem 47

Solve x^2=10 by taking square roots.

Problem 48

Solve 3x^2=75 by taking square roots.

Problem 49

Solve 4x^2=20 by taking square roots.

Problem 50

Solve (x+2)^2=11 by taking square roots.

Problem 51

Solve 5(x-3)^2=45 by taking square roots.

Problem 52

Solve 2(x+7)^2=14 by taking square roots.

Problem 53

Solve x^2=4/9 by taking square roots.

Problem 54

Solve (x-1)^2=1/4 by taking square roots.

transform to square-root equation.
15 problems Warmup Practice Mixed Review Assessment
Problem 55

Solve x^2+4x-12=0 by completing the square.

Problem 56

Solve x^2-2x-8=0 by completing the square.

Problem 57

Solve 2x^2+8x+1=0 by completing the square.

Problem 58

Solve x^2+3x-1=0 by completing the square.

Problem 59

Solve x^2+6x+9=0 by completing the square.

Problem 60

Solve x^2+8x+10=0 by completing the square.

Problem 61

Solve x^2-5x+3=0 by completing the square.

Problem 62

Solve 3x^2+12x-15=0 by completing the square.

Problem 63

Solve 2x^2-4x-1=0 by completing the square.

Open in simulator
Problem 64

Solve 3x^2+9x+2=0 by completing the square.

Problem 65

Solve -x^2+2x+4=0 by completing the square.

Problem 66

Solve x^2-4x+5=0 by completing the square.

Problem 67

Solve x^2-10x+1=0 by completing the square.

Problem 68

Solve 4x^2-6x-1=0 by completing the square.

Problem 69

Solve x^2+12x+30=0 by completing the square.

substitute a, b, c and simplify.
15 problems Warmup Practice Mixed Review Assessment
Problem 70

Solve x^2-5x+6=0 using the quadratic formula.

Problem 71

Solve 2x^2+3x-2=0 using the quadratic formula.

Problem 72

Solve x^2+2x-3=0 using the quadratic formula.

Problem 73

Solve 3x^2-2x-1=0 using the quadratic formula.

Problem 74

Solve x^2+7x+10=0 using the quadratic formula.

Problem 75

Solve 2x^2-x-3=0 using the quadratic formula.

Problem 76

Solve 3x^2+10x+8=0 using the quadratic formula.

Open in simulator
Problem 77

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

Problem 78

Solve 4x^2-8x+3=0 using the quadratic formula.

Problem 79

Solve x^2-7x-18=0 using the quadratic formula.

Problem 80

Solve 5x^2+13x+6=0 using the quadratic formula.

Problem 81

Solve 6x^2-7x+2=0 using the quadratic formula.

Problem 82

Solve x^2-4x-12=0 using the quadratic formula.

Problem 83

Solve 10x^2-11x-6=0 using the quadratic formula.

Problem 84

Solve x^2+8x+15=0 using the quadratic formula.

evaluate `b^2-4ac`.
15 problems Warmup Practice Mixed Review Assessment
Problem 85

Use the discriminant to determine the number and type of solutions for x^2-6x+9=0.

Problem 86

Use the discriminant to determine the number and type of solutions for x^2-5x+4=0.

Problem 87

Use the discriminant to determine the number and type of solutions for x^2-4x+1=0.

Problem 88

Use the discriminant to determine the number and type of solutions for 2x^2+3x+5=0.

Problem 89

Use the discriminant to determine the number and type of solutions for x^2+2x+1=0.

Problem 90

Use the discriminant to determine the number and type of solutions for x^2+5x+6=0.

Problem 91

Use the discriminant to determine the number and type of solutions for x^2+3x+1=0.

Problem 92

Use the discriminant to determine the number and type of solutions for x^2+x+1=0.

Problem 93

Use the discriminant to determine the number and type of solutions for 4x^2-4x+1=0.

Open in simulator
Problem 94

Use the discriminant to determine the number and type of solutions for 3x^2-10x+3=0.

Problem 95

Use the discriminant to determine the number and type of solutions for 2x^2+5x+1=0.

Problem 96

Use the discriminant to determine the number and type of solutions for 3x^2-2x+1=0.

Problem 97

Use the discriminant to determine the number and type of solutions for 9x^2+12x+4=0.

Problem 98

Use the discriminant to determine the number and type of solutions for x^2-x-6=0.

Problem 99

Use the discriminant to determine the number and type of solutions for x^2+6x+2=0.

simplify square roots of negatives using i.
15 problems Warmup Practice Mixed Review Assessment
Problem 100

Solve x^2+9=0 and express complex solutions using i.

Problem 101

Solve x^2+4x+8=0 and express complex solutions using i.

Problem 102

Solve 2x^2+8x+10=0 and express complex solutions using i.

Problem 103

Solve x^2-2x+5=0 and express complex solutions using i.

Problem 104

Solve x^2+1=0 and express complex solutions using i.

Problem 105

Solve x^2+25=0 and express complex solutions using i.

Problem 106

Solve 3x^2+12=0 and express complex solutions using i.

Problem 107

Solve x^2+2x+2=0 and express complex solutions using i.

Open in simulator
Problem 108

Solve x^2-4x+5=0 and express complex solutions using i.

Problem 109

Solve x^2+6x+10=0 and express complex solutions using i.

Problem 110

Solve 2x^2-6x+5=0 and express complex solutions using i.

Problem 111

Solve x^2+x+1=0 and express complex solutions using i.

Problem 112

Solve 3x^2+2x+1=0 and express complex solutions using i.

Problem 113

Solve x^2-6x+13=0 and express complex solutions using i.

Problem 114

Solve 4x^2+1=0 and express complex solutions using i.

identify inspection, factoring, square roots, completing square, or formula.
15 problems Warmup Practice Mixed Review Assessment
Problem 115

Choose the most efficient method to solve (x-2)(x+7)=0.

Problem 116

Choose the most efficient method to solve x^2-81=0.

Problem 117

Choose the most efficient method to solve x^2+6x+5=0.

Problem 118

Choose the most efficient method to solve 3x^2+x-4=0.

Problem 119

Choose the most efficient method to solve (2x+1)(x-3)=0.

Problem 120

Choose the most efficient method to solve x(x+5)=0.

Problem 121

Choose the most efficient method to solve x^2-16=0.

Problem 122

Choose the most efficient method to solve 4x^2-25=0.

Problem 123

Choose the most efficient method to solve x^2-7=0.

Problem 124

Choose the most efficient method to solve x^2-5x+6=0.

Open in simulator
Problem 125

Choose the most efficient method to solve x^2+7x+10=0.

Problem 126

Choose the most efficient method to solve 2x^2+5x+3=0.

Problem 127

Choose the most efficient method to solve x^2+2x+5=0.

Problem 128

Choose the most efficient method to solve x^2+6x+2=0.

Problem 129

Choose the most efficient method to solve (x+3)^2=16.

read zeros from factors.
12 problems Warmup Practice Mixed Review Assessment
Problem 130

Read the solutions directly from factored equation (x-4)(x+2)=0.

Problem 131

Read the solutions directly from factored equation 3(x+5)(x-1)=0.

Problem 132

Read the solutions directly from factored equation (2x-7)(x+3)=0.

Problem 133

Read the solutions directly from factored equation -4(x-6)^2=0.

Problem 134

Read the solutions directly from factored equation (x+1)(x-8)=0.

Open in simulator
Problem 135

Read the solutions directly from factored equation (x-3)(x-9)=0.

Problem 136

Read the solutions directly from factored equation 5(x+2)(x+7)=0.

Problem 137

Read the solutions directly from factored equation (3x+1)(x-5)=0.

Problem 138

Read the solutions directly from factored equation (4x-1)(2x-3)=0.

Problem 139

Read the solutions directly from factored equation 2(x+10)^2=0.

Problem 140

Read the solutions directly from factored equation (x-1)(x+2)(x-3)=0.

Problem 141

Read the solutions directly from factored equation (5x+2)x^2=0.

use vertex-form structure.
15 problems Warmup Practice Mixed Review Assessment
Problem 142

Solve vertex-form equation (x-3)^2-16=0.

Problem 143

Solve vertex-form equation 2(x+1)^2=18.

Problem 144

Solve vertex-form equation -(x-5)^2+9=0.

Problem 145

Solve vertex-form equation 3(x-2)^2+6=0.

Problem 146

Solve vertex-form equation (x+2)^2 - 25 = 0.

Problem 147

Solve vertex-form equation 4(x-1)^2 - 36 = 0.

Problem 148

Solve vertex-form equation -2(x+4)^2 + 8 = 0.

Problem 149

Solve vertex-form equation (x-1)^2 + 4 = 0.

Problem 150

Solve vertex-form equation 2(x+3)^2 + 10 = 0.

Problem 151

Solve vertex-form equation -(x-7)^2 - 16 = 0.

Problem 152

Solve vertex-form equation (x-5)^2 = 0.

Problem 153

Solve vertex-form equation 3(x+6)^2 = 0.

Problem 154

Solve vertex-form equation (1/2)(x-4)^2 - 8 = 0.

Problem 155

Solve vertex-form equation (1/3)(x+2)^2 + 3 = 0.

Open in simulator
Problem 156

Solve vertex-form equation 7x^2 = 0.

solve and reject irrelevant roots.
15 problems Warmup Practice Mixed Review Assessment
Problem 157

Solve the quadratic context and identify the valid solution: A rectangle area is 48. Its length is x+2 and width is x.

Problem 158

Solve the quadratic context and identify the valid solution: A ball height is h=-16t^2+48t. When does it hit the ground after launch?.

Problem 159

Solve the quadratic context and identify the valid solution: A garden area is 84 with dimensions x and x+5.

Problem 160

Solve the quadratic context and identify the valid solution: Revenue is R=-p^2+20p. At what positive price is revenue 96?.

Problem 161

Solve the quadratic context and identify the valid solution: A square has its side increased by 3 units on one side and 2 units on the other, forming a rectangle with area 56. Find the original side length.

Problem 162

Solve the quadratic context and identify the valid solution: A ball is dropped from a height of 64 feet. The height is given by h = -16t^2 + 64. When does it hit the ground?.

Open in simulator
Problem 163

Solve the quadratic context and identify the valid solution: The product of two consecutive positive integers is 72. Find the smaller integer.

Problem 164

Solve the quadratic context and identify the valid solution: A rectangular garden is 10m by 12m. A path of uniform width x is built around it. If the total area of the garden and path is 168 m^2, what is the width of the path?.

Problem 165

Solve the quadratic context and identify the valid solution: The sum of the squares of two consecutive odd positive integers is 130. Find the smaller integer.

Problem 166

Solve the quadratic context and identify the valid solution: A right triangle has legs of length x and x+7. The hypotenuse is 17. Find the length of the shorter leg.

Problem 167

Solve the quadratic context and identify the valid solution: Two pipes fill a tank. Pipe A takes x hours to fill the tank alone. Pipe B takes 3 hours longer than Pipe A. Together they fill the tank in 2 hours. How long does Pipe A take alone?.

Problem 168

Solve the quadratic context and identify the valid solution: A boat travels 24 miles upstream and 24 miles downstream. The speed of the current is 2 mph. If the total trip takes 5 hours, what is the speed of the boat in still water?.

Problem 169

Solve the quadratic context and identify the valid solution: Find a positive number such that if you subtract 5 from it and multiply the result by the number itself, you get 14.

Problem 170

Solve the quadratic context and identify the valid solution: A square piece of cardboard has 2-inch squares cut from each corner. The sides are then folded up to form an open box. If the volume of the box is 72 cubic inches, what was the original side length of the cardboard?.

Problem 171

Solve the quadratic context and identify the valid solution: A company's profit P (in dollars) for selling x units of a product is given by P = -x^2 + 100x - 2400. How many units must be sold to achieve a profit of $100?.

substitute root or use factor relationships.
12 problems Warmup Practice Mixed Review Assessment
Problem 172

Find the missing coefficient in x^2+kx+12=0 given that x=3 is a solution.

Problem 173

Find the missing coefficient in 2x^2+bx-10=0 given that x=2 is a solution.

Problem 174

Find the missing coefficient in x^2-5x+c=0 given that x=1 is a solution.

Problem 175

Find the missing coefficient in ax^2+6x+8=0 given that x=-2 is a solution.

Problem 176

Find the missing coefficient in x^2+7x+c=0 given that x=-4 is a solution.

Problem 177

Find the missing coefficient in 3x^2+bx-6=0 given that x=1 is a solution.

Problem 178

Find the missing coefficient in ax^2-10x+8=0 given that x=2 is a solution.

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Problem 179

Find the missing coefficient in x^2+kx-18=0 given that x=-3 is a solution.

Problem 180

Find the missing coefficient in 2x^2+5x+p=0 given that x=-1/2 is a solution.

Problem 181

Find the missing coefficient in mx^2-x-6=0 given that x=3 is a solution.

Problem 182

Find the missing coefficient in 4x^2+qx+1=0 given that x=-1/2 is a solution.

Problem 183

Find the missing coefficient in x^2-8x+r=0 given that x=5 is a solution.

interpret real roots as zeros.
15 problems Warmup Practice Mixed Review Assessment
Problem 184

Connect the solutions of y=(x-2)(x+5) to the graph's x-intercepts.

Problem 185

Connect the solutions of y=x^2-6x+9 to the graph's x-intercepts.

Problem 186

Connect the solutions of y=x^2+4 to the graph's x-intercepts.

Problem 187

Connect the solutions of y=2(x+1)(x-4) to the graph's x-intercepts.

Problem 188

Connect the solutions of y=(x-3)(x+2) to the graph's x-intercepts.

Problem 189

Connect the solutions of y=(x+4)^2 to the graph's x-intercepts.

Problem 190

Connect the solutions of y=x^2-9 to the graph's x-intercepts.

Problem 191

Connect the solutions of y=3(x-1)(x-5) to the graph's x-intercepts.

Problem 192

Connect the solutions of y=x^2+2x+1 to the graph's x-intercepts.

Problem 193

Connect the solutions of y=x^2-4x+13 to the graph's x-intercepts.

Problem 194

Connect the solutions of y=(x+6)(x-2) to the graph's x-intercepts.

Problem 195

Connect the solutions of y=2x(x-7) to the graph's x-intercepts.

Problem 196

Connect the solutions of y=x^2-16 to the graph's x-intercepts.

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Problem 197

Connect the solutions of y=-(x-3)^2 to the graph's x-intercepts.

Problem 198

Connect the solutions of y=x^2+x+1 to the graph's x-intercepts.

catch factoring, formula, discriminant, square-root, and complex-number mistakes.
15 problems Warmup Practice Mixed Review Assessment
Problem 199

Diagnose and correct the error in this quadratic solution: x^2=25, so x=5 only.

Problem 200

Diagnose and correct the error in this quadratic solution: x^2+5x+6=(x+1)(x+6).

Problem 201

Diagnose and correct the error in this quadratic solution: For x^2+4x+8=0, discriminant is 48.

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Problem 202

Diagnose and correct the error in this quadratic solution: (2x-3)(x+4)=0 gives x=3 and x=-4.

Problem 203

Diagnose and correct the error in this quadratic solution: x^2 - 5x + 6 = (x-1)(x-6).

Problem 204

Diagnose and correct the error in this quadratic solution: For x^2+3x+2=0, x = (3 ± sqrt(9-8))/2.

Problem 205

Diagnose and correct the error in this quadratic solution: For 2x^2+x-3=0, discriminant is 1^2 - 4(2)(-3) = 1 - 24 = -23.

Problem 206

Diagnose and correct the error in this quadratic solution: x^2+6x=10, so x^2+6x+9=10, (x+3)^2=10.

Problem 207

Diagnose and correct the error in this quadratic solution: (x-5)(x+2)=0 gives x=-5 and x=2.

Problem 208

Diagnose and correct the error in this quadratic solution: x^2+2x=8, so x(x+2)=8, which means x=8 or x+2=8 (so x=6).

Problem 209

Diagnose and correct the error in this quadratic solution: x = (6 ± sqrt(20))/2 = 6 ± sqrt(5).

Problem 210

Diagnose and correct the error in this quadratic solution: (x+3)^2 = x^2+9.

Problem 211

Diagnose and correct the error in this quadratic solution: 2x^2=50, so x^2=50, x=±sqrt(50).

Problem 212

Diagnose and correct the error in this quadratic solution: For x^2+x+1=0, discriminant is 1^2-4(1)(1) = -3, so there are no solutions.

Problem 213

Diagnose and correct the error in this quadratic solution: 2x^2+8x=10, so x^2+4x=10, x^2+4x+4=10+4, (x+2)^2=14.