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Week 7 · Chapter

Chapter 7 — Quadratic Equations

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 6 (solve quadratic equations by factoring, the square root property, completing the square, and the quadratic formula; use the discriminant to predict the number of real solutions; and analyze quadratic functions and their graphs, including the vertex, intercepts, and maximum or minimum value) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you solve equations where the unknown number is multiplied by itself, which is called squared (written with a small raised 2, as in x²). Those equations usually have two answers instead of one, and you will learn four ways to find both. You will also learn to read one small number, the discriminant (the part of a formula that tells you how many real answers to expect), before you do any solving. Every answer you find, you will check by putting it back into the original equation. When an equation describes something real, such as a ball landing, you will also decide which answers make sense.

Why you already care about this

Picture an object launched straight up from the edge of an 80-foot platform at 64 feet per second. It climbs, slows, and falls past the platform to the ground. One question matters to anyone standing below: when does it land? Its height after t seconds is −16t² + 64t + 80 feet, so you want the time when that height is 0.

That question has a t² in it, and your Week 2 moves cannot get t alone when it is squared. The same thing happens on the ground. Say you want a patio 3 feet longer than it is wide, covering exactly 40 square feet. The width times the length gives an equation with a squared width in it. A landscaper, a sports analyst tracking a kicked ball, and a shop owner looking for the price where profit hits zero all meet this same kind of equation.

This chapter hands you four tools for it. Each one turns a squared equation back into steps you already trust.

By the end of this chapter you will be able to:

  • write a quadratic equation in standard form and name a, b, and c with their signs;
  • solve by factoring and the zero-product property, without ever dividing by the variable;
  • solve a square set equal to a number with the square root property, keeping both roots;
  • solve by completing the square and by the quadratic formula, and give an exact answer;
  • use the discriminant to predict two real solutions, one repeated solution, or no real solution;
  • solve a quadratic equation in a real setting, reject an answer the setting rules out, and pick a method on purpose.

Carry this sentence with you: set it equal to zero, then factor — never divide by x.

Where we are

  • Already covered: Week 1 set the exponent and sign rules; Week 2 solved linear equations; Weeks 3 to 5 built functions, lines, and systems; Week 6 multiplied and factored polynomials.
  • This week: solving quadratic equations by factoring, the square root property, completing the square, and the quadratic formula, plus the discriminant.
  • How they connect: every factoring move this week is a Week 6 move, and every equation step is a Week 2 move done to both sides.
  • Where we're going next: Week 8, Midterm Review & Exam, reviews Weeks 1 to 7 by the one core move each topic asks for, and then comes the midterm.

Each earlier week supplies a tool you use here. Week 1 showed that (−7)² = (−7)(−7) = 49, because a negative base to an even power is positive. Week 2 gave you the one principle of solving: do the same thing to both sides, then check by substitution. Week 3 introduced the launched object's height function h(t) = −16t² + 64t + 80. Week 6 taught you to factor, so you can already turn x² − 6x + 5 into (x − 1)(x − 5). This week you set that product equal to 0 and solve.

1. Spot a quadratic equation: standard form

Think about a square rug. If each side is x feet long, its area is x times x square feet, which is x². Any equation with a squared unknown, and no higher power, is a quadratic equation.

  • A quadratic equation is an equation in one variable that can be written as ax² + bx + c = 0 with a not equal to zero, so the highest power of the variable is 2.
  • The standard form of a quadratic equation is the arrangement ax² + bx + c = 0 with every term on the left side and 0 on the right, so the coefficients a, b, and c can be read off with their signs.

The number a multiplies x², b multiplies x, and c stands alone. If a were 0, the x² term would vanish and the equation would be a Week 2 linear equation. That is why a ≠ 0 is part of the definition.

Worked example — reading the week's parabola. Find a, b, and c in x² − 6x + 5 = 0.

  1. Check the right side. It is already 0, so the equation is in standard form.
  2. Read a. The x² term has no number written in front, so a = 1.
  3. Read b. The x term is −6x. The minus sign belongs to b, so b = −6.
  4. Read c. The number standing alone is 5, so c = 5.

In words: a = 1, b = −6, and c = 5. You will use these three numbers again and again this week.

Worked example — rearranging first. Put x² = 7x into standard form.

  1. Get 0 on one side. You subtract 7x from both sides: x² − 7x = 0.
  2. Read the coefficients. a = 1, b = −7, and there is no number standing alone, so c = 0.

In words: an equation is in standard form only when one side is 0. You move every term to the left before you read a, b, or c.

Misconception check. Students often read b in x² − 6x + 5 = 0 as 6. It is tempting because the 6 is the number you see, and the minus looks like an operation between terms. The cure: a coefficient carries the sign in front of it. Here b = −6, and that sign will matter in the formula later.

2. Factor and use the zero-product property

Here is a fact you have used without naming it. If two numbers multiply to 0, at least one of them must be 0. No pair of nonzero numbers, however you choose them, multiplies to 0. That one fact turns a factored equation into two easy ones.

  • The zero-product property is the rule that if a product of factors equals 0, then at least one of the factors equals 0, so each factor can be set equal to 0 and solved.

To use it, you need a product on one side and 0 on the other. Week 6 gave you the product: factor the quadratic side.

Worked example — x² − 5x + 6 = 0. Factor, then split.

  1. Find the two numbers. You need a product of 6 and a sum of −5. The pair −2 and −3 works, because (−2)(−3) = 6 and −2 + (−3) = −5.
  2. Write the factors. (x − 2)(x − 3) = 0.
  3. Use the zero-product property. x − 2 = 0 or x − 3 = 0.
  4. Solve each one. x = 2 or x = 3.
  5. Check x = 2. 2² − 5(2) + 6 = 4 − 10 + 6 = 0. It works.
  6. Check x = 3. 3² − 5(3) + 6 = 9 − 15 + 6 = 0. It works.

In words: the equation has two solutions, 2 and 3, and each one makes the left side exactly 0.

Here is how you can hunt for the pair in step 1. List the whole-number pairs that multiply to 6 and add each pair.

Whole-number pairs whose product is 6, with each pair's sum, for factoring x² − 5x + 6
Pair Product Sum
1 and 6 6 7
−1 and −6 6 −7
2 and 3 6 5
−2 and −3 6 −5

Only the last row has the sum −5, so that pair is the one you want.

Worked example — the week's parabola. Solve x² − 6x + 5 = 0 by factoring.

  1. Find the two numbers. You need a product of 5 and a sum of −6. They are −1 and −5.
  2. Write the factors. (x − 1)(x − 5) = 0.
  3. Split and solve. x − 1 = 0 gives x = 1, and x − 5 = 0 gives x = 5.
  4. Check. At x = 1, 1 − 6 + 5 = 0. At x = 5, 25 − 30 + 5 = 0.

In words: x = 1 or x = 5. Those are the two inputs where the table of x² − 6x + 5 shows a 0.

Here is that table. The two zeros sit at x = 1 and x = 5, and the values on either side of x = 3 mirror each other.

Values of the week's parabola x² − 6x + 5 for whole-number inputs from 0 to 6
x 0 1 2 3 4 5 6
x² − 6x + 5 5 0 −3 −4 −3 0 5

In practice (your tools). Open the free Desmos graphing calculator and type y = x^2 - 6x + 5. Click each place where the curve crosses the horizontal axis. Desmos labels the two points (1, 0) and (5, 0), which are the two solutions you found. The graph is a picture of your algebra, so use it as a second opinion, never as a replacement for the check.

Misconception check. Students often read the factor (x − 2) and write x = −2. It is tempting because the 2 has a minus sign in front of it. The cure: set the factor equal to 0 and solve it, x − 2 = 0, so x = 2. Your check catches the slip at once, since (−2)² − 5(−2) + 6 = 4 + 10 + 6 = 20, not 0.

3. Keep every answer: never divide by the variable

Imagine a friend says, "x² = 7x, so divide both sides by x, and x = 7." It feels efficient. But try x = 0 in the original equation: 0² = 0 and 7 · 0 = 0. So 0 is a solution too, and dividing by x threw it away. Dividing by x quietly assumes x is not 0, and that assumption was never checked.

The cure is the move from Week 6: pull out the greatest common factor. Then the zero-product property hands you every answer.

Worked example — 3x² − 12x = 0. Factor out the common factor, then split.

  1. Find what both terms share. 3x² and 12x both contain 3 and x, so the common factor is 3x.
  2. Factor it out. 3x(x − 4) = 0.
  3. Split. 3x = 0 gives x = 0, and x − 4 = 0 gives x = 4.
  4. Check x = 0. 3(0)² − 12(0) = 0 − 0 = 0.
  5. Check x = 4. 3(4)² − 12(4) = 3 · 16 − 48 = 48 − 48 = 0.

In words: both 0 and 4 solve the equation. Dividing 3x² − 12x = 0 by 3x leaves x − 4 = 0, so only x = 4 survives, and the answer x = 0 is lost.

Worked example — x² = 7x, done safely. You subtract 7x to get x² − 7x = 0. You factor out x to get x(x − 7) = 0. So x = 0 or x = 7. Check x = 7: 7² = 49 and 7 · 7 = 49. Check x = 0: 0² = 0 and 7 · 0 = 0.

Misconception check. Students often divide both sides by x because it is the move that worked in Week 2. In Week 2 you divided by numbers, never by an expression that could be 0. Honesty note: dividing by x is allowed only when you already know x is not 0, and in a solve you do not know that yet. The cure: move everything to one side and factor.

4. Undo a square: the square root property

Picture a square rug that covers 49 square feet. How long is each side? You need a number that times itself is 49, and 7 works. But (−7)² is also 49. A rug cannot have a negative side, but a bare equation does not know about rugs. So the equation x² = 49 has two answers.

  • The square root property is the rule that if a square equals a nonnegative number k, then the quantity being squared equals the positive or the negative square root of k, which gives two answers when k is positive.
  • The plus-or-minus sign is the symbol ± placed between two numbers to name both their sum and their difference at once, so 3 ± 2 means 5 and 1. Written in front of one number, as in ±7, it names that number and its negative.

The square root symbol √ always means the principal square root, the nonnegative one, so √49 = 7. The ± is what brings back the negative answer.

Worked example — x² = 49. Take the square root of both sides.

  1. Apply the property. x = ±√49.
  2. Evaluate the root. √49 = 7, so x = ±7, which means x = 7 or x = −7.
  3. Check. 7² = 49, and (−7)² = (−7)(−7) = 49.

In words: the equation has two solutions, 7 and −7. If x stood for the rug's side, only 7 would make sense, and you would reject −7 for that reason.

Worked example — a shifted square, (x − 3)² = 16. Here the thing being squared is x − 3, not x.

  1. Apply the property to the whole square. x − 3 = ±4, because √16 = 4.
  2. Split the two cases. x − 3 = 4 or x − 3 = −4.
  3. Solve each. x = 3 + 4 = 7, or x = 3 − 4 = −1.
  4. Check x = 7. (7 − 3)² = 4² = 16.
  5. Check x = −1. (−1 − 3)² = (−4)² = 16.

In words: x = 7 or x = −1. Both answers land 4 units from 3 on the number line, one on each side.

Misconception check. Students often write x² = 49 and stop at x = 7. It is tempting because the square root key on a calculator gives only 7. The cure: whenever you take the square root of both sides of an equation, write ± first. Then decide, from the setting, whether both answers stay.

5. Build a square: completing the square

The square root property is fast, but only when one side is already a square. Most equations are not. Completing the square is a way to make one. Think of tiles: an x-by-x square, some x-by-1 strips, and a few 1-by-1 tiles. If the strips are split evenly between two sides of the big square, a small corner is left empty. Filling that corner completes the square.

  • Completing the square is a solving method that adds the square of half the x-coefficient to both sides of x² + bx = (a number), turning the left side into a perfect square such as (x − 3)².

The recipe is three moves: move c to the right side, take half of b, square it, and add that square to both sides.

Worked example — the week's parabola, a second way. Solve x² − 6x + 5 = 0 by completing the square.

  1. Move c. You subtract 5 from both sides: x² − 6x = −5.
  2. Half of b. Half of −6 is −3.
  3. Square it. (−3)² = 9.
  4. Add 9 to both sides. x² − 6x + 9 = −5 + 9 = 4.
  5. Write the square. x² − 6x + 9 = (x − 3)², so (x − 3)² = 4.
  6. Square root property. x − 3 = ±2.
  7. Solve. x = 3 + 2 = 5, or x = 3 − 2 = 1.

In words: x = 1 or x = 5, the same two answers factoring gave in Section 2. Two methods, one answer. That agreement is a check in itself.

Worked example — x² + 6x + 5 = 0. This time b is positive.

  1. Move c. x² + 6x = −5.
  2. Half of b, squared. Half of 6 is 3, and 3² = 9.
  3. Add 9 to both sides. x² + 6x + 9 = 4, so (x + 3)² = 4.
  4. Square root property. x + 3 = ±2.
  5. Solve. x = −3 + 2 = −1, or x = −3 − 2 = −5.
  6. Check x = −1. (−1)² + 6(−1) + 5 = 1 − 6 + 5 = 0.
  7. Check x = −5. (−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0.

In words: x = −1 or x = −5. Both answers check.

Misconception check. Students often add the 9 to the left side only. It is tempting because the 9 is "for" the left side, to finish its square. But an equation is a balance, and adding 9 to one side breaks it. The cure: say "both sides" out loud at step 4, as you did in Week 2.

6. Use the quadratic formula on any quadratic

Completing the square works on every quadratic equation. If you complete the square on ax² + bx + c = 0 once, with letters instead of numbers, you get a formula that does the job in one step. You do not need to rebuild it each time; you use it.

  • The quadratic formula is x = (−b ± √(b² − 4ac)) ⁄ (2a), which gives every real solution of ax² + bx + c = 0 from its three coefficients.

Worked example — the week's parabola, a third way. Solve x² − 6x + 5 = 0 with the formula.

  1. Read the coefficients. a = 1, b = −6, c = 5.
  2. Find the part under the root. (−6)² − 4(1)(5) = 36 − 20 = 16.
  3. Take its root. √16 = 4.
  4. Fill in the formula. x = (−(−6) ± 4) ⁄ (2 · 1) = (6 ± 4) ⁄ 2.
  5. Split the ±. x = 10 ⁄ 2 = 5, or x = 2 ⁄ 2 = 1.

In words: x = 1 or x = 5. One equation, three methods, one answer.

Worked example — an equation that will not factor. Solve x² − 4x + 1 = 0.

  1. Read the coefficients. a = 1, b = −4, c = 1. So −b = −(−4) = 4.
  2. Find the part under the root. (−4)² − 4(1)(1) = 16 − 4 = 12.
  3. Simplify the root. √12 = √(4 · 3) = 2√3, because 4 is a perfect square with √4 = 2.
  4. Fill in the formula. x = (4 ± 2√3) ⁄ 2.
  5. Divide both terms on top by 2. x = 2 ± √3.
  6. Give a decimal as a check. 2 + √3 ≈ 3.732 and 2 − √3 ≈ 0.268.

In words: the exact answers are 2 + √3 and 2 − √3. The decimals are rounded, so they are a check on size, not the answer itself.

You can check 2 + √3 exactly with the Week 6 square of a sum. (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3, and 4(2 + √3) = 8 + 4√3. So (7 + 4√3) − (8 + 4√3) + 1 = 7 − 8 + 1 = 0. The √3 parts cancel, and the rest is 0.

In practice (your tools). In Desmos, type y = x^2 - 4x + 1 and click the two places where the curve crosses the horizontal axis. Desmos shows them as about 0.268 and 3.732. Your exact answer 2 ± √3 matches the picture, and the picture alone could never have told you the √3.

Misconception check — two classic slips. Students often write −b as −4 when b = −4. That gives (−4 ± 2√3) ⁄ 2 = −2 ± √3, which is wrong. The cure: −b is the opposite of b, and the opposite of −4 is 4. Students also divide only the radical by 2, which gives 4 ± √3, which is also wrong. The cure: the fraction bar sits under the whole top, so both 4 and 2√3 get divided by 2.

7. Read the discriminant before you solve

Look back at the formula. Everything depends on the number under the square root. If it is positive, the ± gives two different answers. If it is 0, adding and subtracting 0 gives the same number twice. If it is negative, no real number has that square root. That one number is worth a name.

  • The discriminant is the quantity b² − 4ac, the part under the square root in the quadratic formula, whose sign tells you before you solve whether a quadratic equation has two real solutions, one repeated solution, or no real solution.
  • A repeated solution is the single answer a quadratic equation has when its discriminant is 0, called repeated because the formula's plus and minus branches land on the same number.
  • No real solution is the outcome when a quadratic equation's discriminant is negative: no real number makes the equation true, because no real number squares to a negative number.

Here are the three cases, side by side.

The sign of the discriminant b² − 4ac and the number of real solutions it predicts, with one example of each
Sign of b² − 4ac Number of real solutions Example Discriminant
Positive two x² − 6x + 5 = 0 16
Zero one repeated solution 2x² − 4x + 2 = 0 0
Negative none x² + 2x + 5 = 0 −16

Worked example — a zero discriminant. Solve 2x² − 4x + 2 = 0.

  1. Read the coefficients. a = 2, b = −4, c = 2.
  2. Find the discriminant. (−4)² − 4(2)(2) = 16 − 16 = 0.
  3. Predict. A discriminant of 0 means one repeated solution.
  4. Solve. x = (4 ± 0) ⁄ 4 = 1.
  5. Check. 2(1)² − 4(1) + 2 = 2 − 4 + 2 = 0.

In words: the only solution is x = 1.

Worked example — a negative discriminant. Solve x² + 2x + 5 = 0.

  1. Read the coefficients. a = 1, b = 2, c = 5.
  2. Find the discriminant. 2² − 4(1)(5) = 4 − 20 = −16.
  3. Decide. −16 is negative, and no real number squares to −16. So there is no real solution.

In words: you write "no real solution" and stop. Caveat: mathematicians have a larger system, the complex numbers, in which this equation does have solutions. That system is real mathematics, and it is outside this course.

For the week's parabola, the discriminant is 16, and 16 > 0, so it has two real solutions, 1 and 5. The discriminant told you "two" before you found them.

Misconception check. Students often "solve" x² + 2x + 5 = 0 anyway, writing √(−16) = −4. It is tempting because 4 and −16 both contain a 16. But (−4)² = 16, not −16, so −4 is not a square root of −16. The cure: when the discriminant is negative, you write the words "no real solution", not a number.

8. Apply the methods and choose one on purpose

Back to the launched object from the opening. Its height in feet t seconds after launch is h(t) = −16t² + 64t + 80. The 16 is half of gravity's pull of 32 feet per second per second, in a model that ignores air resistance. At t = 0, 1, 2, 3, 4, and 5 seconds the heights are 80, 128, 144, 128, 80, and 0 feet.

Worked example — when does it land? Solve −16t² + 64t + 80 = 0.

  1. Make the numbers smaller. You divide every term by −16: t² − 4t − 5 = 0.
  2. Factor. You need a product of −5 and a sum of −4: −5 and 1. So (t − 5)(t + 1) = 0.
  3. Split. t = 5 or t = −1.
  4. Ask the setting. t = −1 is one second before the launch, so it is rejected.
  5. Check t = 5. −16(5)² + 64(5) + 80 = −16(25) + 320 + 80 = −400 + 320 + 80 = 0.

In words: the object lands 5 seconds after launch. The equation's other answer is real algebra, but it does not describe this flight.

Worked example — the patio. A rectangular patio is 3 feet longer than it is wide, and its area is 40 square feet.

  1. Name the unknown. Let w be the width in feet. Then the length is w + 3.
  2. Write the equation. Width times length is area: w(w + 3) = 40.
  3. Distribute. w² + 3w = 40.
  4. Get 0 on one side. You subtract 40: w² + 3w − 40 = 0.
  5. Factor. You need a product of −40 and a sum of 3: 8 and −5. So (w + 8)(w − 5) = 0.
  6. Split and judge. w = −8 or w = 5. A width cannot be negative, so w = 5.
  7. Finish and check. The length is 5 + 3 = 8 feet, and 5 × 8 = 40 square feet.

In words: the patio is 5 feet wide and 8 feet long.

Choosing a method. All four methods give the same answers when they apply, so you choose by speed and safety. Here is a guide.

Which of the four solving methods to try first for each shape of quadratic equation, with this week's example of each
What the equation looks like Method to try first This week's example
Standard form, and the numbers factor at a glance factoring and the zero-product property x² − 5x + 6 = 0
One square set equal to a number the square root property (x − 3)² = 16
Standard form, and no whole-number factors appear the quadratic formula x² − 4x + 1 = 0
Any quadratic, when you want to see why the formula works completing the square x² + 6x + 5 = 0

In practice (your tools). In Desmos, type y = -16x^2 + 64x + 80. Desmos uses x for the horizontal axis, so x plays the role of t here. Click where the curve meets the horizontal axis on the right, and Desmos labels (5, 0). The curve also meets the axis at (−1, 0), the answer you rejected, off to the left of the launch.

Misconception check. Students often report both answers, t = 5 and t = −1, as "the times it lands." It is tempting because the algebra produced both. The cure: an answer to an equation is not yet an answer to the question. Ask what the variable means. A time before launch, or a negative width, is ruled out by the setting, versus ruled out by the algebra.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

  • (a) Think of a moment from your own life when something was thrown, kicked, or dropped. Describe what a quadratic equation could tell you about it, and say which answer you might have to reject and why.
  • (b) Pick the method from this chapter you feel most sure of and the one you feel least sure of. For each, name one equation from this chapter where it is the fastest choice, and say what makes it fast there.

Key terms

  • Quadratic equation — An equation in one variable that can be written as ax² + bx + c = 0 with a not equal to zero, so the highest power of the variable is 2. (first used: 1. Spot a quadratic equation: standard form)
  • Standard form of a quadratic equation — The arrangement ax² + bx + c = 0 with every term on the left side and 0 on the right, so the coefficients a, b, and c can be read off with their signs. (first used: 1. Spot a quadratic equation: standard form)
  • Zero-product property — The rule that if a product of factors equals 0, then at least one of the factors equals 0, so each factor can be set equal to 0 and solved. (first used: 2. Factor and use the zero-product property)
  • Square root property — The rule that if a square equals a nonnegative number k, then the quantity being squared equals the positive or the negative square root of k, which gives two answers when k is positive. (first used: 4. Undo a square: the square root property)
  • Plus-or-minus sign — The symbol ± placed between two numbers to name both their sum and their difference at once, so 3 ± 2 means 5 and 1. (first used: 4. Undo a square: the square root property)
  • Completing the square — A solving method that adds the square of half the x-coefficient to both sides of x² + bx = (a number), turning the left side into a perfect square such as (x − 3)². (first used: 5. Build a square: completing the square)
  • Quadratic formula — The formula x = (−b ± √(b² − 4ac)) ⁄ (2a), which gives every real solution of ax² + bx + c = 0 from its three coefficients. (first used: 6. Use the quadratic formula on any quadratic)
  • Discriminant — The quantity b² − 4ac, the part under the square root in the quadratic formula, whose sign tells you before you solve whether a quadratic equation has two real solutions, one repeated solution, or no real solution. (first used: 7. Read the discriminant before you solve)
  • Repeated solution — The single answer a quadratic equation has when its discriminant is 0, called repeated because the formula's plus and minus branches land on the same number. (first used: 7. Read the discriminant before you solve)
  • No real solution — The outcome when a quadratic equation's discriminant is negative: no real number makes the equation true, because no real number squares to a negative number. (first used: 7. Read the discriminant before you solve)

Summary

  • A quadratic equation has a squared unknown; put it in standard form, with 0 on one side, before you solve.
  • Each coefficient keeps its sign: in x² − 6x + 5 = 0, b is negative 6.
  • Factor, then set each factor equal to 0; that is the zero-product property.
  • Never divide both sides by the variable; factor out the common factor instead.
  • Taking a square root of both sides gives a plus-or-minus pair of answers.
  • Completing the square adds the square of half of b to both sides.
  • The quadratic formula solves every quadratic; divide the whole top by 2a.
  • The discriminant's sign predicts two, one, or no real solutions.
  • Check every answer by substitution, and reject any answer the setting rules out.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. Solve x² + x − 12 = 0 by factoring, and check both answers.
  2. Solve 2x² = 10x without losing an answer.
  3. A square mural for a community center will cover 81 square feet. Write an equation for its side length s, solve it with the square root property, and say which answer the mural keeps.
  4. Without solving, use the discriminant to say how many real solutions each equation has: (a) 3x² − 2x + 4 = 0; (b) x² − 10x + 25 = 0. Then find the solution of (b).

Answers (worked reasoning — read after attempting)

  1. You need a product of −12 and a sum of 1, which is 4 and −3. So (x + 4)(x − 3) = 0, and x = −4 or x = 3. Check x = −4: (−4)² + (−4) − 12 = 16 − 4 − 12 = 0. Check x = 3: 3² + 3 − 12 = 9 + 3 − 12 = 0. Reading (x + 4) as x = 4 is the sign slip from Section 2; the check would give 16 + 4 − 12 = 8, not 0.
  2. Subtract 10x from both sides: 2x² − 10x = 0. Factor out the common factor 2x: 2x(x − 5) = 0. So x = 0 or x = 5. Check x = 0: 2(0)² = 0 and 10(0) = 0. Check x = 5: 2(25) = 50 and 10(5) = 50. Dividing both sides by 2x would have left only x = 5 and lost x = 0.
  3. The area of a square is its side times itself, so s² = 81. Then s = ±√81 = ±9. Check: 9² = 81 and (−9)² = 81. A side length cannot be negative, so the mural keeps s = 9 feet and rejects −9.
  4. (a) a = 3, b = −2, c = 4, so the discriminant is (−2)² − 4(3)(4) = 4 − 48 = −44. It is negative, so there is no real solution. (b) a = 1, b = −10, c = 25, so the discriminant is (−10)² − 4(1)(25) = 100 − 100 = 0. That means one repeated solution: x = (10 ± 0) ⁄ 2 = 5. Check: 5² − 10(5) + 25 = 25 − 50 + 25 = 0.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the midterm review practices choosing a method for every topic from Weeks 1 to 7.

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