Restoration Series · Quiz 40 · Two Answers

The Guide

Before you begin

A curve crossing a line at two points — which is what a quadratic equation looks like, and why it usually has two answers rather than one.

This quiz finishes family 12, and it needs one thing from you before you start: you have to be able to factor. If Quiz 39 is not done, do it. If it is done but slow, do a couple of sections of Algebra Practice 2 first. The central method here is factoring, and there is no way around that.

There are two halves. The first is about equations that are still linear but no longer simple — variables on both sides, and parentheses in the way. The second is quadratic equations, which is the last big idea in GED algebra.

Variables on both sides

Everything you know about solving equations still holds: whatever you do to one side you do to the other, and the aim is to end with the variable alone on one side.

What is new is that the variable is now in two places, so there is a step before the usual ones: get all the variable terms onto one side and all the plain numbers onto the other.

5x + 3 = 2x + 18

Take 2x from both sides. The right side loses its x entirely:

3x + 3 = 18

And now it is the two-step equation from The Balance. Take 3 from both sides, then divide by 3:

3x = 15, so x = 5

Check by putting 5 back into the original: the left is 25 + 3 = 28, the right is 10 + 18 = 28. They match, so it is right.

Which side to move. It does not matter mathematically — you will get the same answer either way. It matters for comfort: move the smaller variable term, and you keep the one you are left with positive. In 5x + 3 = 2x + 18, moving the 2x leaves 3x. Moving the 5x instead leaves −3x = −15, which is the same answer through one more negative than you needed.

Parentheses in the way

When an equation has terms grouped in parentheses, multiply them out first, then solve as usual. That is the whole rule.

3(x + 4) = 2(x + 9)

Multiply out both sides — the number outside reaches every term inside:

3x + 12 = 2x + 18

Now it is the kind of equation you just solved. Take 2x from both sides, then 12:

x + 12 = 18, so x = 6

Watch for a minus in front of parentheses, because it behaves here exactly as it did in Quiz 39: it changes the sign of every term inside.

20 − 2(x − 3) = 10 becomes 20 − 2x + 6 = 10, which is 26 − 2x = 10, so 2x = 16 and x = 8.

The +6 is where people lose this. −2 times −3 is +6, not −6.

What makes an equation quadratic

Every equation so far has had the variable to the first power only. A quadratic equation has an x² in it, and that one change alters what an answer even looks like.

The standard shape, which is on the formula sheet, is

ax² + bx + c = 0

where a, b and c are numbers and a is not zero. The whole of this half of the quiz is about getting an equation into that shape and then finding the values of x that make it true.

Here is the part that surprises people. A linear equation has one answer. A quadratic usually has two, and both of them are correct.

The picture at the start of the Guide is why. A quadratic drawn on a graph is a U-shaped curve called a parabola, and solving the equation means finding where that curve crosses the horizontal line. A U crosses a line twice.

Solving a quadratic means finding where the curve meets the line. A U meets a line twice, which is where the two answers come from.

Solving by factoring

The method rests on one fact about numbers, and it is worth saying on its own because everything else follows from it.

If two things multiply to zero, at least one of them is zero. There is no other way to get zero from a multiplication. If you are told P × Q = 12, that tells you very little — it could be 3 and 4, or 2 and 6, or 24 and ½. But if you are told P × Q = 0, then either P is zero or Q is zero, and there is no third option. Zero is the only number that does this, and that is why the next step insists on getting everything onto one side.

So the method has four steps.

One. Get everything onto one side so the other side is 0.

Two. Factor the side that has everything on it.

Three. Set each binomial equal to zero, separately. (A binomial is an expression with exactly two terms, such as x + 3. Factoring a quadratic like the ones here gives two of them, multiplied together, and Quiz 39 is where they come from.)

Four. Solve those two small equations. Their answers are the two answers.

Take x² + 8x + 15 = 0. It is already equal to zero, so step one is done. Factoring it is the work of Quiz 39 — two numbers multiplying to 15 and adding to 8, which are 3 and 5:

(x + 3)(x + 5) = 0

Two things multiplying to zero, so one of them is zero:

x + 3 = 0 or x + 5 = 0

x = −3 or x = −5

Both are answers. Check them both, which is quick: (−3)² + 8(−3) + 15 = 9 − 24 + 15 = 0, and (−5)² + 8(−5) + 15 = 25 − 40 + 15 = 0.

The sign flips. The binomials read (x + 3) and (x + 5), and the answers are −3 and −5. This catches people constantly. The binomial is not the answer; the number that makes the binomial zero is the answer, and to make x + 3 into zero you need x to be −3. So the answers are the opposites of the numbers in the binomials.

When it is not already equal to zero

Most questions do not hand it to you in standard shape. Step one is the one that gets skipped.

x² + 5x = 24

You cannot factor this and read off answers, because the right side is not zero. Move the 24 across first:

x² + 5x − 24 = 0

Now factor: two numbers multiplying to −24 and adding to 5 are 8 and −3.

(x + 8)(x − 3) = 0, so x = −8 or x = 3

And if the equation has parentheses, multiply out before moving anything. (x + 2)(x − 3) = 6 is not solved by setting the binomials to 6 and 1. Multiply out to get x² − x − 6 = 6, then move the 6 to get x² − x − 12 = 0, which factors to (x − 4)(x + 3) = 0, giving x = 4 or x = −3.

Two shortcuts worth having

When there is no plain number at the end, the common factor is an x:

x² − 7x = 0 factors as x(x − 7) = 0, so x = 0 or x = 7.

Zero is a perfectly good answer and it is the one people forget, because it feels like nothing. It is not nothing; it is a number that makes the equation true.

When there is no middle term, it is a difference of squares:

x² − 49 = 0 factors as (x + 7)(x − 7) = 0, so x = 7 or x = −7.

The quadratic formula

Some quadratics do not factor with whole numbers. There is nothing wrong with them — they have answers, the answers are just not tidy. For those there is a formula, and it is printed on the sheet they give you:

It works on every quadratic, including the ones that factor. The price is arithmetic, so use it when factoring has failed rather than instead of trying.

To use it, get the equation into ax² + bx + c = 0 first — the same step one as before — then read off the three numbers with their signs and substitute.

Take x² + 5x + 6 = 0, which does factor, so that the formula can be checked against an answer we can get another way. Here a = 1, b = 5, c = 6.

The part under the square root first, because it is where the work is: b² − 4ac = 5² − 4(1)(6) = 25 − 24 = 1. The square root of 1 is 1.

Then the rest: x = (−5 ± 1) ÷ 2.

The ± sign means do it twice, once adding and once subtracting, and that is where the two answers come from.

Adding: (−5 + 1) ÷ 2 = −4 ÷ 2 = −2. Subtracting: (−5 − 1) ÷ 2 = −6 ÷ 2 = −3.

And factoring gives the same pair: x² + 5x + 6 = (x + 2)(x + 3) = 0, so x = −2 or x = −3. Two roads, one answer.

Using the calculator on this. The formula is where the TI-30XS earns its place. Work out b² − 4ac on its own first and write it down; then take its square root; then do the two divisions separately. Trying to type the whole formula in one line is how parentheses get lost. And mind the sign of b: the formula starts with minus b, so if b is already negative that becomes a plus.

The guard

Get one side to zero before factoring. The method depends on it entirely and it is the step most often skipped.

Expect two answers, and give both unless the question says otherwise.

The answers are the opposites of the numbers in the binomials. (x − 4) gives x = 4; (x + 4) gives x = −4.

Check by substituting. Both answers, back into the original equation. As with factoring, this is a check that cannot mislead you.

Read a, b and c with their signs. In x² − 3x − 10 = 0, b is −3 and c is −10, not 3 and 10.

Worked Examples

Three to study before you start

Example 1 · Variables on both sides, with parentheses

Solve 4(x − 1) = 2x + 10.

Multiply out the parentheses first: 4 times x is 4x, and 4 times −1 is −4.

4x − 4 = 2x + 10

Now move the smaller variable term. Take 2x from both sides: 2x − 4 = 10. Add 4 to both sides: 2x = 14. Divide by 2: x = 7.

Check in the original, not in your working: the left is 4(7 − 1) = 4 × 6 = 24, and the right is 14 + 10 = 24. Checking in the original catches a mistake made in the very first step, which checking in your working cannot do.

Example 2 · A quadratic that has to be moved first

Solve x² = 3x + 28.

Step one, everything to one side. Take 3x and 28 from both sides:

x² − 3x − 28 = 0

Step two, factor. Two numbers multiplying to −28 and adding to −3: the pairs for 28 are 1 and 28, 2 and 14, 4 and 7, and 7 − 4 = 3, so with the larger one negative it is −7 and +4.

(x − 7)(x + 4) = 0

Steps three and four, each binomial to zero: x = 7 or x = −4.

Check both in the original. For 7: 49 on the left, 21 + 28 = 49 on the right. For −4: 16 on the left, −12 + 28 = 16 on the right. Both work, and both are answers.

Example 3 · The formula, on one that does not factor

Solve x² + 4x − 6 = 0, giving the answers to one decimal place.

Try factoring first: two numbers multiplying to −6 and adding to 4. The pairs for 6 are 1 and 6, and 2 and 3. Their differences are 5 and 1. Neither is 4, so it does not factor with whole numbers. Use the formula.

a = 1, b = 4, c = −6.

Under the root: b² − 4ac = 16 − 4(1)(−6) = 16 + 24 = 40. Note the sign: subtracting a negative added.

√40 is about 6.32.

Adding: (−4 + 6.32) ÷ 2 = 2.32 ÷ 2 = 1.2. Subtracting: (−4 − 6.32) ÷ 2 = −10.32 ÷ 2 = −5.2.

A rough check that the answers are sensible: 1.2² + 4(1.2) − 6 = 1.44 + 4.8 − 6 = 0.24, near enough to zero given the rounding. If it had come out at 12 rather than 0.24 you would know something had gone wrong.

The Quiz · Ten Questions

Now you

Work on paper. A calculator is fine for question 9. Give both answers to every quadratic, and check each one by substituting it back into the original equation. Questions 6 and 10 have two parts.

Solve: 7x − 4 = 3x + 16
Solve: 5(x + 2) = 3(x + 8)
Solve by factoring: x² + 7x + 12 = 0
Solve by factoring: x² − 2x − 15 = 0
- A) x = 3 or x = −5
- B) x = −3 or x = 5
- C) x = −3 or x = −5
- D) x = 3 or x = 5
Solve: x² + 3x = 10
Solve each of these. (a) x² − 6x = 0 (b) x² − 81 = 0

Reading A reading rest stop, for the stubborn ones.

Ten minutes with the last section of this page — The Company, below the answer key — before you finish the quiz. It is about the clay tablets that show people were solving these exact equations three and a half thousand years before anyone wrote x, and about who those people were and what they were counting. It will not help you with question 7. It may change what you think you are doing here.

No photograph at this rest stop. The drawings in this quiz are all above.

Solve: (x + 3)(x − 2) = 6
Ravi solves x² + 2x = 8 by factoring the left side into x(x + 2) and then writing x = 8 or x + 2 = 8. Is he right? Say what went wrong and solve it properly.
Use the quadratic formula to solve x² − 6x + 4 = 0. Give both answers to one decimal place.
A rectangular garden is 3 meters longer than it is wide, and its area is 40 square meters. (a) If the width is x, write an equation for the area and put it in the form ax² + bx + c = 0. (b) Solve it, and say what the width of the garden is.

Answer Key

Check your work

1 x = 5

Take 3x from both sides: 4x − 4 = 16. Add 4: 4x = 20. Divide by 4: x = 5. Check in the original: the left is 35 − 4 = 31 and the right is 15 + 16 = 31. Moving the 3x rather than the 7x is what keeps the numbers positive.

2 x = 7

Multiply out both sides first: 5x + 10 = 3x + 24. Take 3x: 2x + 10 = 24. Take 10: 2x = 14. So x = 7. Check: 5(9) = 45 and 3(15) = 45. The most common slip here is multiplying only the first term inside each set of parentheses and getting 5x + 2 = 3x + 8.

3 x = −3 or x = −4

Two numbers multiplying to 12 and adding to 7 are 3 and 4, so (x + 3)(x + 4) = 0. Each binomial to zero gives x = −3 and x = −4. Both signs flip, which is the thing to keep watching. Check the first: 9 − 21 + 12 = 0.

4 B) x = −3 or x = 5

Two numbers multiplying to −15 and adding to −2: the pair is 3 and 5 with the larger one negative, so −5 and +3. That gives (x − 5)(x + 3) = 0 and the answers 5 and −3. A) is the same two numbers with both signs the wrong way around, which is what you get by reading the binomials as the answers instead of flipping them. C) and D) both keep the signs matching, which cannot be right when the last term is negative. Substituting is the fastest way to settle it: 25 − 10 − 15 = 0 for x = 5, and 9 + 6 − 15 = 0 for x = −3.

5 x = 2 or x = −5

Move the 10 across first — this is the step the question is testing: x² + 3x − 10 = 0. Then two numbers multiplying to −10 and adding to 3 are 5 and −2, giving (x + 5)(x − 2) = 0 and the answers −5 and 2. If you factored the left side as it stood into x(x + 3) and set the pieces equal to 10, that is question 8's mistake, and it does not work because the fact about multiplying to zero is about zero and nothing else.

6 (a) x = 0 or x = 6 (b) x = 9 or x = −9

(a) There is no plain number, so the common factor is x: x(x − 6) = 0. The two factors are x and (x − 6), so x = 0 or x = 6. Zero is a real answer and the one usually dropped — check it if you doubt it: 0 − 0 = 0. Dividing both sides by x at the start would have given x = 6 alone and quietly lost the other answer, which is why you factor rather than divide. (b) No middle term and two squares subtracted: (x + 9)(x − 9) = 0, so x = 9 or x = −9.

7 x = 3 or x = −4

The two binomials multiply to 6, not to zero, so nothing can be read off them yet. Multiply out: x² + x − 6 = 6. Move the 6 across: x² + x − 12 = 0. Factor: two numbers multiplying to −12 and adding to 1 are 4 and −3, so (x + 4)(x − 3) = 0, giving x = −4 or x = 3. The tempting wrong move is setting x + 3 = 6 and x − 2 = 1, which gives 3 and 3 — and 3 does happen to be one of the real answers, which makes the wrong method look as though it worked. It did not; it lost the other answer and was right by accident.

8 No. The right side is 8, not 0, so nothing can be read off the factors. Moving the 8 first gives x = 2 or x = −4.

Ravi factored correctly — x² + 2x really is x(x + 2) — and then used a rule that does not exist. If two things multiply to 8, that tells you nothing about either one on its own: they could be 2 and 4, or 1 and 8, or 16 and a half. Only zero forces one of the factors to be zero, and that is the whole reason for step one. Done properly: x² + 2x − 8 = 0, which factors to (x + 4)(x − 2) = 0, giving x = −4 or x = 2. Ravi's method would have given 8 and 6, and neither works: 64 + 16 is 80, not 8.

9 x = 5.2 or x = 0.8

Here a = 1, b = −6, c = 4. Under the root: (−6)² − 4(1)(4) = 36 − 16 = 20, and √20 is about 4.47. The formula begins with −b, and b is already negative, so −b is +6. Adding: (6 + 4.47) ÷ 2 = 10.47 ÷ 2 = 5.2. Subtracting: (6 − 4.47) ÷ 2 = 1.53 ÷ 2 = 0.8. Try factoring first if you like — two whole numbers multiplying to 4 and adding to −6 would have to be −1 and −4, or −2 and −2, and neither adds to −6 — which is how you know the formula is needed.

10 (a) x² + 3x − 40 = 0 (b) x = 5 or x = −8; the width is 5 meters

(a) If the width is x, the length is x + 3, and the area is width times length: x(x + 3) = 40. Multiply out and move the 40 across: x² + 3x − 40 = 0. (b) Two numbers multiplying to −40 and adding to 3 are 8 and −5, so (x + 8)(x − 5) = 0 and x = −8 or x = 5. Both are answers to the equation, but only one is an answer to the question: a garden cannot be −8 meters wide. So the width is 5, the length is 8, and 5 × 8 = 40, which checks. This is the one place in the family where you throw an answer away, and the reason is not mathematical — the equation is perfectly happy with −8. It is that the equation is standing in for a garden, and gardens have positive widths. Word problems will often do this, so after solving, read the question again and ask whether both answers make sense as the thing they describe.

Reading your results

Eight to ten: that is family 12, and with it the algebra on this test. Take a practice test. Five to seven: sort the misses. If 1, 2 or 7 went wrong, the fault is in the setting-up rather than the solving, so redo them writing every step on its own line. If 3, 4, 5 or 6 went wrong, the fault is the factoring itself, and the cure is Algebra Practice 2 rather than more quadratics. Under five: go back to Quiz 39. Nothing in this quiz is hard once factoring is quick, and nothing in it is possible while factoring is slow.

The Company · An Interlude

The clay that knew the answer

Around 1800 BC, in the cities of southern Mesopotamia, a student pressed a reed into wet clay and worked out a problem we would now write as a quadratic equation.

We know this because the clay survived. Thousands of tablets have been dug out of the ground in what is now Iraq, and among the accounts and letters and contracts are mathematical ones: problem texts, tables, and the working-out of scribes in training. They are the oldest algebra we have, and they are about fifteen hundred years older than the Greeks and two and a half thousand years older than the word algebra.

One tablet type poses the problem like this: I have added the area and the side of a square, and the result is such-and-such. Find the side. In our symbols that is x² + x = a number — which is exactly question 5 of this quiz.

They had no symbols. No x, no equals sign, no plus or minus, none of it — all of that is between two and four thousand years in their future. The problems are written out in words, in Akkadian, and the method is given as a recipe: halve this, square it, add it to that, take the square root, subtract the half you started with. Follow the recipe and the answer comes out.

And the recipe is the quadratic formula. Not something like it — it is the same sequence of operations, in the same order, that you would get by working the formula through by hand. Four thousand years ago, in a language nobody has spoken for millennia, written with a cut reed on river mud.

Two things about this are worth sitting with.

The first is what they were counting. These are not abstract puzzles, or not only that. Mesopotamian mathematics grew up around the things a city state had to keep track of: how much grain is in the store, how many days of labor a canal will take, what is owed on a loan and at what interest, how to divide an inheritance, how much earth comes out of a ditch of given length and depth. The quadratics turn up in problems about fields — a rectangle whose length is so much more than its width, with a stated area — which is question 10 of this quiz, more or less word for word, three and a half thousand years early.

The second is who did it. The tablets came out of scribal schools, and a scribe was a salaried professional: someone trained for years to do the counting that a temple, a palace or a merchant house needed done. Not philosophers. Bookkeepers, surveyors, administrators — skilled workers whose job was to make the numbers come out. Some of the tablets are plainly exercises, copied and re-copied by students, with the same problem worked again with different figures, which is the oldest practice sheet in the world.

A few of those students signed their work. Most did not.

Why this is in a GED quiz. There is a story people are told about mathematics, which is that it descends from a small number of geniuses and arrives in a classroom as something to be received. The clay says otherwise. This particular piece of mathematics was worked out by people doing a job, because the job required it, and it has been passed hand to hand ever since by people who mostly did not get their names on it. You are not being let in on a secret. You are joining the end of a very long line of people who needed to know how wide the field was.

Sources: the Old Babylonian mathematical tablets, of which the best known are in the collections of Yale, Columbia and the British Museum; the standard modern accounts of their contents and of the scribal schools they came from.

Where this goes. That is family 12 complete, and with it every algebra family on the test: Family 11 for expressions, Family 12 for equations, Family 13 for lines and slope. For practice, Algebra Practice 2 has 366 problems on the multiplying and factoring that all of this runs on.