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Week 8 · Chapter

Chapter 8 — Midterm Review & Exam

College Algebra Free sample edition

The module's primary reading · ~35–45 minutes · Objective 1 (classify real numbers, and simplify and evaluate algebraic expressions using the properties of real numbers, the order of operations, and the rules of integer exponents), Objective 2 (solve linear equations and linear inequalities in one variable, including absolute-value equations and inequalities, and express solution sets in interval notation), Objective 3 (use function notation; determine the domain and range of a function; evaluate functions at numeric and algebraic inputs; and form sums, differences, products, quotients, and compositions of functions), Objective 4 (analyze linear functions (slope, intercepts, and the forms of a line, including parallel and perpendicular lines) and solve systems of linear equations and inequalities in two variables), Objective 5 (add, subtract, and multiply polynomials, apply the special products, and factor polynomial expressions by the greatest common factor, trinomial patterns, the difference of squares, and grouping), 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 early this week, before the practice set and the discussion and well before the midterm.

In plain language

This week you look back at the first seven weeks and see that every problem asked you for one kind of work. We call that work the problem's core move (the one job a problem asks you to do, such as simplify, solve, or factor). You also learn one new habit, choosing a method (reading a problem and naming its move before you write a step). The clue you read is the problem cue (a feature of how the problem is written, such as an equals sign or a squared variable). The midterm covers the same seven weeks, and the habit carries you through the second half of the course.

Why you already care about this

Think about the last time you opened a toolbox with 30 tools in it. A loose screw needs 1 of them, a screwdriver. A stuck nail needs a hammer, and a wobbly shelf needs a level. You do not try all 30. You look at the problem for a moment, and the problem tells you which tool to grab.

Algebra works the same way. In seven weeks you have collected six toolboxes: simplify, solve, functions, lines and systems, polynomials, and quadratic equations. Each one holds three or four tools. The hard part of a mixed test is rarely the tool itself. It is picking up the right one fast, without guessing.

This chapter walks the six toolboxes once more, with one fresh problem for each, and then teaches the habit of reading a problem before you touch it. Every answer comes with its check, because a check is what turns a guess into a result.

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

  • simplify an expression with the order of operations and the exponent rules, keeping −2³ and (−3)² apart;
  • solve a linear equation, a linear inequality, and an absolute-value equation, and check each answer;
  • evaluate a function at a negative input, find the domain of a square-root function, and evaluate a composition;
  • find a slope from two points, write the line's equation, and solve a system of two equations;
  • multiply a binomial square without dropping the middle term, and factor a trinomial;
  • solve one quadratic equation by four methods and predict its number of real solutions;
  • read a mixed problem's cue and name its core move before you start.

Carry this sentence with you: a cue is not a step — read it before you write.

Where we are

  • Already covered: Weeks 1 and 2 set the rules and solved equations; Weeks 3 to 5 built functions, lines, and systems; Weeks 6 and 7 factored polynomials and solved quadratics.
  • This week: a review of all seven weeks, organized by the one core move each topic asks for, plus one new habit, choosing a method, and the midterm.
  • How they connect: every core move from the first half becomes a step inside a bigger problem in the second half, so naming the move is the skill that ties them together.
  • Where we're going next: Week 9, Quadratic Functions & Their Graphs, turns this week's parabola into a picture and asks what its formula says about its graph.

Each week handed the next one a tool. Week 1's exponent rules let Week 6 multiply polynomials. Week 2's solving moves finish every system in Week 5 and every factored quadratic in Week 7. Week 3's function notation is how Week 4 names a line. Use the sections below as a checklist. If a step feels shaky, go back to that week's chapter and read its worked example again.

1. Simplify: follow the order and the exponent rules

Picture a receipt with no total at the bottom. Nobody asks you to find a mystery number. You are only asked to tidy what is there into one clean value. That is simplifying: a problem with no equals sign asks you to simplify or evaluate.

Two Week 1 rules do the work:

  • Order of operations: grouping symbols first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. An exponent binds before a leading minus sign, so −2³ means −(2³).
  • Exponent rules: the power of a product raises every factor, and the power rule multiplies the exponents, so (x²)³ = x⁶.

Worked example — a negated power and a negative base. Simplify −2³ + (−3)².

  1. The first term: the exponent binds first, so −2³ = −(2³) = −8.
  2. The second term: the parentheses make −3 the base, so (−3)² = (−3)(−3) = 9.
  3. Add: −8 + 9 = 1.

In words: one minus sign sits outside its power and one sits inside. If you read (−3)² as −3², you get −9, and then −8 + (−9) = −17, which is wrong.

Worked example — a power of a product. Simplify (2x²y)³.

  1. Raise every factor: (2x²y)³ = 2³ · (x²)³ · y³.
  2. Work each piece: 2³ = 8, and (x²)³ = x⁶ because the exponents multiply (2 × 3 = 6).
  3. Put it together: 8 · x⁶ · y³ = 8x⁶y³.
  4. Check at x = 2 and y = 1: the original gives (2 · 2² · 1)³ = (2 · 4)³ = 8³ = 512, and the answer gives 8 · 2⁶ · 1³ = 8 · 64 = 512.

Both sides give 512, so you can trust the result. The trap answer 8x⁵y³ gives 8 · 2⁵ = 8 · 32 = 256, which does not match. The true value of 512 versus the trap's 256 is a 2-fold gap, so a quick check catches the slip.

Misconception check — adding when you should multiply. Students often think (x²)³ is x⁵, because the product rule taught them to add exponents. It is tempting because both rules involve two exponents. But the product rule is for two powers multiplied side by side, like x² · x³. A power raised to a power multiplies. The cure is a quick number check: put in a small value such as 2 and compare.

In practice (your tools). In Desmos, type -2^3 + (-3)^2 on an empty line, and it shows 1. Type -3^2 on the next line, and it shows −9, which is the trap in action. Desmos follows the order of operations exactly, so it checks your hand work; it does not replace it.

Memory hook: the minus outside waits its turn; the minus inside gets squared.

2. Solve one-variable statements: equation, inequality, absolute value

Imagine two bike shops on the same street. Shop A lets you ride the first hour free and then charges 4 dollars for each hour after that. Shop B charges a 6-dollar fee plus 2 dollars an hour. Which is cheaper? An equals sign turns that question into a number you can find.

Three cues tell you which solving tool to use:

  • An equals sign with x to the first power asks for the one number that makes the statement true. You do the same thing to both sides.
  • An inequality sign asks for a whole set of numbers. The flip rule applies: when you multiply or divide both sides by a negative number, the sign reverses.
  • Absolute-value bars set equal to a positive number call for the split rule: |X| = a with a positive gives X = a or X = −a.

Worked example — the bike shops. For x hours, shop A costs 4(x − 1) dollars and shop B costs 2x + 6 dollars. Solve 4(x − 1) = 2x + 6.

  1. Distribute: 4x − 4 = 2x + 6.
  2. Collect the x terms: subtract 2x from both sides to get 2x − 4 = 6.
  3. Isolate: add 4 to get 2x = 10, then divide by 2 to get x = 5.
  4. Check x = 5: the left side is 4(5 − 1) = 4 · 4 = 16, and the right side is 2 · 5 + 6 = 10 + 6 = 16.

In words: at 5 hours both shops cost 16 dollars. For a shorter or longer ride, one shop is cheaper, and the equation tells you where the switch happens.

Worked example — an inequality with a negative divisor. Solve −3x + 2 ≤ 11.

  1. Subtract 2 from both sides: −3x ≤ 9.
  2. Divide both sides by −3 and flip the sign: x ≥ −3.
  3. Write the interval: [−3, ∞). The bracket says −3 is included, and ∞ always gets a parenthesis.
  4. Check: x = 0 gives −3 · 0 + 2 = 2, and 2 ≤ 11 is true. The endpoint x = −3 gives −3 · (−3) + 2 = 9 + 2 = 11, and 11 ≤ 11 is true.

Worked example — the split rule. Solve |2x − 3| = 7.

  1. Split: 2x − 3 = 7 or 2x − 3 = −7.
  2. Solve the first: 2x = 10, so x = 5.
  3. Solve the second: 2x = −4, so x = −2.
  4. Check both: |2 · 5 − 3| = |7| = 7, and |2 · (−2) − 3| = |−7| = 7.

Misconception check — the forgotten flip. Students often divide −3x ≤ 9 by −3 and write x ≤ −3. It is tempting because every other step of solving keeps the sign the same. Test x = −4 in the original: −3 · (−4) + 2 = 12 + 2 = 14, and 14 ≤ 11 is false. So x ≤ −3 is wrong. The cure: whenever you divide by a negative number, say "flip" out loud.

Memory hook: an equation finds a number; an inequality finds a set.

3. Work with functions: evaluate, find a domain, compose

A vending machine is a function. You press one button, and you get exactly one snack. The button is the input, and the snack is the output. Function notation writes that as f(x), read "f of x", and it means one output, not f times x.

Three function moves came up in Week 3:

  • To evaluate: replace every x with the input, wrapped in parentheses, and follow the order of operations.
  • To find a domain: a square root needs its radicand to be zero or more, and a denominator cannot be zero.
  • To compose: (p ∘ q)(x) = p(q(x)), read "p of q of x", so you work inside first.

Worked example — evaluate at a negative input. For f(x) = x² − 2x, find f(−3).

  1. Substitute with parentheses: f(−3) = (−3)² − 2(−3).
  2. Work each term: (−3)² = 9, and −2(−3) = +6.
  3. Add: 9 + 6 = 15.

So f(−3) = 15. If you write (−3)² as −3², you get −9 + 6 = −3, which is wrong.

Worked example — the domain of a radical. Find the domain of √(x − 4).

  1. Set the radicand to zero or more: x − 4 ≥ 0.
  2. Solve: x ≥ 4.
  3. Write the interval: [4, ∞).

Test the edge: at x = 4 the value is √0 = 0, which is a real number, so you include 4 with a bracket. At x = 3 the radicand is −1, and no real number squares to −1, so 3 is left out.

Worked example — inside first. With p(x) = x² and q(x) = x + 1:

  1. (p ∘ q)(2) = p(q(2)) = p(3) = 9.
  2. (q ∘ p)(2) = q(p(2)) = q(4) = 5.

Order matters: the same two functions give 9 one way and 5 the other.

Misconception check — the open endpoint. Students often write the domain of √(x − 4) as (4, ∞), leaving 4 out. It is tempting because "the radicand must be positive" sounds right. But √0 = 0 is a perfectly good output. The rule is "zero or more", so 4 belongs in the domain and gets a bracket.

In practice (your tools). In Desmos, type f(x) = x^2 - 2x on one line and f(-3) on the next; Desmos shows 15. Then type y = sqrt(x - 4) and notice the graph starts at the point (4, 0) and runs to the right, which is the domain [4, ∞) drawn as a picture.

Memory hook: parentheses around every input — then work inside first.

4. Read lines and solve systems of two equations

Picture a cool morning that warms up steadily. One hour before noon the thermometer reads 2 °C. Three hours after noon it reads 10 °C. How fast is it warming, and what did it read at noon? Two points and a steady rate are the cue for a line.

From Weeks 4 and 5:

  • The slope, which measures how steep a line is, for the line through two points is m = (y₂ − y₁) ⁄ (x₂ − x₁): rise over run, with the y-values on top.
  • Point-slope form, y − y₁ = m(x − x₁), builds the line from one point and the slope; slope-intercept form, y = mx + b, then shows the y-intercept b.
  • Two equations with two unknowns form a system, and its solution is an ordered pair that makes both equations true.

Worked example — the warming morning. Let x be the hours after noon, so one hour before noon is x = −1. The two readings are the points (−1, 2) and (3, 10).

  1. Slope: m = (10 − 2) ⁄ (3 − (−1)) = 8 ⁄ 4 = 2. The temperature rises 2 °C per hour.
  2. Point-slope form with (−1, 2): y − 2 = 2(x + 1).
  3. Distribute and solve for y: y − 2 = 2x + 2, so y = 2x + 4.
  4. Read the intercept: at x = 0, y = 4, so it was 4 °C at noon.
  5. Check (3, 10): 2 · 3 + 4 = 6 + 4 = 10.

Worked example — a workday split two ways. A 7-hour workday is split between two tasks, and one task takes 1 hour longer than the other. With x and y as the hours on each task, x + y = 7 and x − y = 1.

  1. Add the equations: the y and −y cancel, so 2x = 8 and x = 4.
  2. Back-substitute: 4 + y = 7, so y = 3.
  3. Check (4, 3) in both: 4 + 3 = 7 is true, and 4 − 3 = 1 is true.

In words: one task takes 4 hours and the other takes 3.

Misconception check — losing the double negative. Students often compute the run as 3 − 1 instead of 3 − (−1). It is tempting because the minus sign in front of the 1 is easy to drop. That mistake gives 8 ⁄ 2 = 4, which is wrong. Putting the run on top gives 4 ⁄ 8 = 1⁄2, also wrong. The cure: write the subtraction with the parentheses, 3 − (−1) = 4, every time.

In practice (your tools). In Desmos, type (-1,2) and (3,10) on two lines to plot the points, then type y = 2x + 4. The line passes through both points. For the system, type x + y = 7 and x - y = 1, and click where the two lines cross to see the point (4, 3).

Memory hook: y's on top; one pair, both equations.

5. Multiply and factor polynomials

Imagine a square rug that is 2x + 3 feet on each side. How many square feet of floor does it cover? Multiplying the side by itself gives the area, and the result is a polynomial. A product of two binomials, or a binomial squared, is the cue to multiply. The word "factor" is the cue to run that work backward.

From Week 6:

  • The square of a sum follows the special product (a + b)² = a² + 2ab + b². The middle term 2ab is the piece students drop.
  • Factoring undoes multiplication. For a trinomial x² + bx + c, you look for two numbers whose product is c and whose sum is b, and you check the factors by expanding them.

Worked example — the rug. Expand (2x + 3)².

  1. Name the pieces: a = 2x and b = 3.
  2. Apply the pattern: (2x)² + 2(2x)(3) + 3².
  3. Work each term: 4x² + 12x + 9.
  4. Check at x = 1: (2 · 1 + 3)² = 5² = 25, and 4 + 12 + 9 = 25.

So the rug covers 4x² + 12x + 9 square feet. The trap 4x² + 9 gives 4 + 9 = 13 at x = 1, which does not match.

Worked example — factor back. Factor x² − 6x + 5.

  1. Find the pair: you need a product of 5 and a sum of −6. The numbers −1 and −5 work.
  2. Write the factors: x² − 6x + 5 = (x − 1)(x − 5).
  3. Check by expanding: x² − 5x − x + 5 = x² − 6x + 5.

Misconception check — squaring each piece. Students often think (2x + 3)² is 4x² + 9. It is tempting because squaring each piece feels like the power-of-a-product rule from Section 1. But that rule is for a product, and 2x + 3 is a sum. A square means the binomial times itself, and the two cross products add up to the middle term 12x.

Memory hook: a square of a sum always has a middle.

6. Solve a quadratic equation four ways, discriminant first

Suppose a friend gives you a puzzle: "I am thinking of a number. Its square, minus 6 times the number, plus 5, equals zero." A squared variable set equal to something is the cue for a quadratic equation. The week's parabola, x² − 6x + 5 = 0, is that puzzle.

From Week 7:

  • Standard form is ax² + bx + c = 0 with a ≠ 0. Here a = 1, b = −6, and c = 5.
  • The discriminant is b² − 4ac. A positive value means two real solutions, zero means one repeated solution, and a negative value means no real solution.
  • The four methods are factoring with the zero-product property, the square root property, completing the square, and the quadratic formula x = (−b ± √(b² − 4ac)) ⁄ (2a).

Worked example — the discriminant first. For x² − 6x + 5 = 0, the discriminant is (−6)² − 4(1)(5) = 36 − 20 = 16. Since 16 is positive, you expect two real solutions before you solve.

Worked example — three paths, one answer. Solve x² − 6x + 5 = 0 three ways.

  1. Factoring: (x − 1)(x − 5) = 0, so x = 1 or x = 5.
  2. Completing the square: move c to get x² − 6x = −5. Half of −6 is −3, and (−3)² = 9. Add 9 to both sides: x² − 6x + 9 = −5 + 9 = 4, which is (x − 3)² = 4.
  3. The square root property finishes it: x − 3 = ±2, so x = 5 or x = 1.
  4. The quadratic formula: x = (6 ± 4) ⁄ 2, so x = 10 ⁄ 2 = 5 or x = 2 ⁄ 2 = 1.
  5. Check: at x = 1, 1 − 6 + 5 = 0. At x = 5, 25 − 30 + 5 = 0.

Every path lands on 1 and 5, which is why you can always use one method to check another. Caveat: the discriminant tells you how many real solutions to expect, not what they are, so you still have to solve.

Worked example — when the discriminant says stop. x² + 2x + 5 = 0 has discriminant 2² − 4(1)(5) = 4 − 20 = −16. No real number squares to a negative number, so you write "no real solution" and stop. In contrast, 2x² − 4x + 2 = 0 has discriminant (−4)² − 4(2)(2) = 16 − 16 = 0, so it has one repeated solution, x = 1.

Misconception check — dividing only part of the top. Students often compute the formula's top, 6 ± 4, and then divide only the 4 by 2, which gives 6 ± 2. It is tempting because the radical looks like its own piece. But the fraction bar covers the whole top. (6 ± 4) ⁄ 2 means 10 ⁄ 2 and 2 ⁄ 2, which give 5 and 1.

In practice (your tools). In Desmos, type y = x^2 - 6x + 5 and click the two places where the curve meets the x-axis; Desmos labels them (1, 0) and (5, 0). Next week you will learn what the rest of that curve means.

Memory hook: read the discriminant before you pick up a pencil.

7. Choose a method: name the core move before you start

Picture a mechanic who listens to an engine for a few seconds before opening the hood. That pause is not wasted time. It decides which tool comes out first. On a mixed test, your first few seconds should do the same job.

Choosing a method has three steps:

  1. Read the problem cue. Look at how the problem is written before you look at its numbers.
  2. Name the core move in a word or two: simplify, solve, split, evaluate, domain, slope, system, multiply, factor, or quadratic.
  3. Then start, and finish with a check.

Here is how the cues line up with the moves from Sections 1 to 6.

Problem cues from Weeks 1 to 7 and the core move each one signals
What you see in the problem The core move
No equals sign Simplify or evaluate
An equals sign, x to the first power Solve a linear equation
An inequality sign Solve, flip on a negative divisor, write an interval
Absolute-value bars equal to a number Split into two equations
A function name around a number Evaluate, with parentheses
A radical or a denominator, asking "which inputs?" Find the domain
Two points Find the slope, then the line
Two equations, two unknowns Solve a system
A product or square of binomials Multiply, keeping the middle term
The word "factor" GCF first, then a pattern
A squared variable in an equation Solve a quadratic equation

For a quadratic equation, a second look at its form picks the method:

  • A square equal to a number, such as x² = 49, calls for the square root property: x = ±7. Check: 7² = 49 and (−7)² = 49.
  • No constant term, such as 3x² − 12x = 0, calls for the greatest common factor: 3x(x − 4) = 0, so x = 0 or x = 4.
  • Whole-number factors that are easy to spot, such as x² − 6x + 5 = 0, call for factoring.
  • A discriminant that is not a perfect square, such as x² − 4x + 1 = 0 with discriminant (−4)² − 4(1)(1) = 16 − 4 = 12, calls for the formula: x = (4 ± √12) ⁄ 2 = 2 ± √3.

Worked example — the launched object. An object launched from a platform lands when −16t² + 64t + 80 = 0, where t is the time in seconds.

  1. Read the cue: a squared variable in an equation, so the move is a quadratic solve.
  2. Read the form: every coefficient divides by −16, so you divide first: t² − 4t − 5 = 0.
  3. Factor: (t − 5)(t + 1) = 0, so t = 5 or t = −1.
  4. Check the setting: t = −1 is before the launch, so you reject it. The object lands at t = 5 seconds.
  5. Check the arithmetic: −16(5)² + 64(5) + 80 = −400 + 320 + 80 = 0.

Worked example — sort a mixed set. Here are six of this week's problems with their moves named first:

  1. |2x − 3| = 7 is a split.
  2. −3x + 2 ≤ 11 is a solve with a possible flip.
  3. f(−3) is an evaluation.
  4. √(x − 4) with "which inputs?" is a domain.
  5. x + y = 7 with x − y = 1 is a system.
  6. x² − 6x + 5 = 0 is a quadratic solve.

Misconception check — dividing by the variable. Students often see 3x² − 12x = 0 and divide both sides by x to "simplify". It is tempting because it makes the equation look like a Week 2 problem. But dividing by x throws away the answer x = 0. Naming the move first prevents it: no constant term means "factor out the GCF", and the zero-product property keeps both answers.

Pause and think. Choose one of these to answer in a sentence or two, in your notes:

  • (a) Which cue in the table do you read fastest, and which one do you most often miss? Name one problem from Weeks 1 to 7 where a missed cue sent you down the wrong path.
  • (b) Pick two moves from the table that look alike to you, such as "multiply" and "factor". Write how you will tell them apart on the midterm.

Honesty note: naming the move does not do the algebra for you. It only makes sure your effort goes into the right algebra, and the check at the end is still what tells you the work holds up.

Memory hook: cue, move, then pencil.

Key terms

  • Core move — The one kind of work a problem asks for, named before any step is written: simplify, solve, evaluate, find a domain, find a slope, solve a system, multiply, factor, or solve a quadratic equation. (first used: 1. Simplify: follow the order and the exponent rules)
  • Problem cue — A feature of a problem's written form, such as an equals sign, an inequality sign, absolute-value bars, a radical, two given points, or a squared variable, that signals which core move the problem asks for. (first used: 2. Solve one-variable statements: equation, inequality, absolute value)
  • Check by substitution — Replacing the variable in the original equation or inequality with a proposed answer and evaluating each side to confirm the statement is true. (first used: 2. Solve one-variable statements: equation, inequality, absolute value)
  • 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: 6. Solve a quadratic equation four ways, discriminant first)
  • Choosing a method — The habit of reading a problem's cue, naming its core move, and picking a procedure before starting, so the first step is chosen on purpose rather than by guessing. (first used: 7. Choose a method: name the core move before you start)

Summary

  • Every problem asks for one core move; name it before you write a step.
  • No equals sign means simplify; an equals sign or an inequality sign means solve.
  • An exponent binds before a leading minus sign, and a power of a power multiplies.
  • Dividing an inequality by a negative number flips the sign.
  • Evaluate with parentheses around the input, and a square root's domain includes the point where the radicand is zero.
  • Slope is rise over run with the y-values on top, and a system's pair must check in both equations.
  • The square of a sum always has a middle term.
  • Read the discriminant first; then let the equation's form pick the quadratic method.
  • Every answer ends with a check by substitution.

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. Simplify −5² + (−4)², and then simplify (3a³b²)².
  2. Name the core move, then solve 5 − 2x > 13. Write the answer in interval notation and check one value.
  3. A smoothie uses 12 ounces of fruit and yogurt in all, with 4 more ounces of fruit than yogurt. Name the core move, then find both amounts and check them.
  4. For each equation, name the method its form calls for, then solve it or explain why it has no real solution: (a) (x + 2)² = 25; (b) x² + 3x + 4 = 0.

Answers (worked reasoning — read after attempting)

  1. The core move is simplify, because there is no equals sign. The exponent binds before the leading minus, so −5² = −(5²) = −25. The parentheses make −4 the base, so (−4)² = 16. Then −25 + 16 = −9. For the second expression, raise every factor: (3a³b²)² = 3² · (a³)² · (b²)² = 9a⁶b⁴, because the exponents multiply (3 × 2 = 6 and 2 × 2 = 4). Adding them instead would give 9a⁵b⁴, the trap from Section 1.
  2. The core move is solve an inequality. Subtract 5 from both sides: −2x > 8. Divide by −2 and flip the sign: x < −4, written (−∞, −4). Check x = −5: 5 − 2(−5) = 5 + 10 = 15, and 15 > 13 is true. The endpoint −4 gives 5 + 8 = 13, and 13 > 13 is false, so −4 gets a parenthesis.
  3. Two unknowns and two conditions make a system. With x ounces of fruit and y ounces of yogurt, x + y = 12 and x − y = 4. Add the equations: 2x = 16, so x = 8. Then 8 + y = 12, so y = 4. Check: 8 + 4 = 12 and 8 − 4 = 4. The smoothie uses 8 ounces of fruit and 4 ounces of yogurt.
  4. (a) A square equals a number, so use the square root property: x + 2 = ±5. Then x = −2 + 5 = 3 or x = −2 − 5 = −7. Check: (3 + 2)² = 25 and (−7 + 2)² = (−5)² = 25. (b) Nothing factors at a glance, so read the discriminant first: 3² − 4(1)(4) = 9 − 16 = −7. It is negative, so the equation has no real solution, and you stop there.

Next up in this module: the lecture, the slide deck, the Readings & Resources page, the practice set, the discussion, and the midterm — and next week a new rule, the vertex formula, reads this parabola's graph.

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