U2 6 solve quadratics by completing the square answers

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0.7 Quadratic Equations 83

0.7 Quadratic Equations

In Section 0.6.1, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that section were carefully constructed so that the polynomials could be factored using the integers. To demonstrate just how contrived the equations had to be, we can solve 2x2 + 5x �3 = 0 by factoring, (2x �1)(x + 3) = 0, from which we obtain x = 12 and x = �3. If we change the 5 to a 6 and try to solve 2x

2 + 6x � 3 = 0, however, we find that this polynomial doesn’t factor over the integers and we are stuck. It turns out that there are two real number solutions to this equation, but they are irrational numbers, and our aim in this section is to review the techniques which allow us to find these solutions.1 In this section, we focus our attention on quadratic equations.

Definition 0.15. An equation is said to be quadratic in a variable X if it can be written in the form AX 2 + BX + C = 0 where A, B and C are expressions which do not involve X and A 6= 0.

Think of quadratic equations as equations that are one degree up from linear equations - instead of the highest power of X being just X = X 1, it’s X 2. The simplest class of quadratic equations to solve are the ones in which B = 0. In that case, we have the following.

Solving Quadratic Equations by Extracting Square Roots If c is a real number with c � 0, the solutions to X 2 = c are X = ±

p c.

Note: If c < 0, X 2 = c has no real number solutions.

There are a couple different ways to see why Extracting Square Roots works, both of which are demonstrated by solving the equation x2 = 3. If we follow the procedure outlined in the previous section, we subtract 3 from both sides to get x2 � 3 = 0 and we now try to factor x2 � 3. As mentioned in the remarks following Definition 0.14, we could think of x2�3 = x2� (

p 3)2 and apply

the Difference of Squares formula to factor x2�3 = (x� p

3)(x + p

3). We solve (x� p

3)(x + p

3) = 0 by using the Zero Product Property as before by setting each factor equal to zero: x �

p 3 = 0

and x + p

3 � 0. We get the answers x = ± p

3. In general, if c � 0, then p

c is a real number, so x2 � c = x2 � (

p c)2 = (x �

p c)(x +

p c). Replacing the ‘3’ with ‘c’ in the above discussion gives

the general result. Another way to view this result is to visualize ‘taking the square root’ of both sides: since x2 = c,p

x2 = p

c. How do we simplify p

x2? We have to exercise a bit of caution here. Note that p

(5)2 and

p

(�5)2 both simplify to p

25 = 5. In both cases, p

x2 returned a positive number, since the negative in �5 was ‘squared away’ before we took the square root. In other words,

p x2 is x if x

is positive, or, if x is negative, we make x positive - that is, p

x2 = |x |, the absolute value of x . So from x2 = 3, we ‘take the square root’ of both sides of the equation to get

p x2 =

p 3. This

simplifies to |x | = p

3, which by Theorem 0.3 is equivalent to x = p

3 or x = � p

3. Replacing the ‘3’ in the previous argument with ‘c,’ gives the general result.

1While our discussion in this section departs from factoring, we’ll see in Chapter 3 that the same correspondence between factoring and solving equations holds whether or not the polynomial factors over the integers.

84 Prerequisites

As you might expect, Extracting Square Roots can be applied to more complicated equations. Consider the equation below. We can solve it by Extracting Square Roots provided we first isolate the perfect square quantity:

2 ✓

x + 3 2

◆2 � 15

2 = 0

2 ✓

x + 3 2

◆2 =

15 2

Add 15 2

✓

x + 3 2

◆2 =

15 4

Divide by 2

x + 3 2

= ± r

15 4

Extract Square Roots

x + 3 2

= ± p

15 2

Property of Radicals

x = �3 2 ±

p 15 2

Subtract 3 2

x = �3 ± p

15 2

Add fractions

Let’s return to the equation 2x2 + 6x � 3 = 0 from the beginning of the section. We leave it to the reader to show that

2 ✓

x + 3 2

◆2 � 15

2 = 2x2 + 6x � 3.

(Hint: Expand the left side.) In other words, we can solve 2x2 + 6x � 3 = 0 by transforming into an equivalent equation. This process, you may recall, is called ‘Completing the Square.’ We’ll revisit Completing the Square in Section 2.3 in more generality and for a different purpose but for now we revisit the steps needed to complete the square to solve a quadratic equation.

Solving Quadratic Equations: Completing the Square To solve a quadratic equation AX 2 + BX + C = 0 by Completing the Square:

1. Subtract the constant C from both sides.

2. Divide both sides by A, the coefficient of X 2. (Remember: A 6= 0.)

3. Add ⇣

B 2A

⌘2 to both sides of the equation. (That’s half the coefficient of X , squared.)

4. Factor the left hand side of the equation as ⇣

X + B2A ⌘2

.

5. Extract Square Roots.

6. Subtract B2A from both sides.

0.7 Quadratic Equations 85

To refresh our memories, we apply this method to solve 3x2 � 24x + 5 = 0:

3x2 � 24x + 5 = 0 3x2 � 24x = �5 Subtract C = 5

x2 � 8x = �5 3

Divide by A = 3

x2 � 8x + 16 = �5 3

+ 16 Add ⇣

B 2A

⌘2 = (�4)2 = 16

(x � 4)2 = 43 3

Factor: Perfect Square Trinomial

x � 4 = ± r

43 3

Extract Square Roots

x = 4 ± r

43 3

Add 4

At this point, we use properties of fractions and radicals to ‘rationalize’ the denominator:2 r

43 3

= r

43 · 3 3 · 3 =

p 129p

9 = p

129 3

We can now get a common (integer) denominator which yields:

x = 4 ± r

43 3

= 4 ± p

129 3

= 12 ±

p 129

3

The key to Completing the Square is that the procedure always produces a perfect square trino- mial. To see why this works every single time, we start with AX 2 + BX + C = 0 and follow the procedure:

AX 2 + BX + C = 0 AX 2 + BX = �C Subtract C

X 2 + BX A

= �C A

Divide by A 6= 0

X 2 + BX A

+ ✓

B 2A

◆2 = �C

A + ✓

B 2A

◆2 Add

✓

B 2A

◆2

(Hold onto the line above for a moment.) Here’s the heart of the method - we need to show that

X 2 + BX A

+ ✓

B 2A

◆2 = ✓

X + B 2A

◆2

To show this, we start with the right side of the equation and apply the Perfect Square Formula from Theorem 0.7

✓

X + B 2A

◆2 = X 2 + 2

✓

B 2A

◆

X + ✓

B 2A

◆2 = X 2 +

BX A

+ ✓

B 2A

◆2 X

2Recall that this means we want to get a denominator with rational (more specifically, integer) numbers.

86 Prerequisites

With just a few more steps we can solve the general equation AX 2 + BX + C = 0 so let’s pick up the story where we left off. (The line on the previous page we told you to hold on to.)

X 2 + BX A

+ ✓

B 2A

◆2 = �C

A + ✓

B 2A

◆2

✓

X + B 2A

◆2 = �C

A +

B2

4A2 Factor: Perfect Square Trinomial

✓

X + B 2A

◆2 = �4AC

4A2 +

B2

4A2 Get a common denominator

✓

X + B 2A

◆2 =

B2 � 4AC 4A2

Add fractions

X + B 2A

= ± r

B2 � 4AC 4A2

Extract Square Roots

X + B 2A

= ± p

B2 � 4AC 2A

Properties of Radicals

X = � B 2A

± p

B2 � 4AC 2A

Subtract B 2A

X = �B ±

p B2 � 4AC 2A

Add fractions.

Lo and behold, we have derived the legendary Quadratic Formula!

Theorem 0.9. Quadratic Formula: The solution to AX 2 + BX + C = 0 with A 6= 0 is:

X = �B ±

p B2 � 4AC 2A

We can check our earlier solutions to 2x2 + 6x � 3 = 0 and 3x2 � 24x + 5 = 0 using the Quadratic Formula. For 2x2 + 6x � 3 = 0, we identify A = 2, B = 6 and C = �3. The quadratic formula gives:

x = �6 ±

p

62 � 4(2)(�3) 2(2)

� �6 ± p

36 + 24 4

= �6 ±

p 60

4

Using properties of radicals ( p

60 = 2 p

15), this reduces to 2(�3± p

15) 4 =

�3± p

15 2 . We leave it to the

reader to show these two answers are the same as �3± p

15 2 , as required.

3

For 3x2 � 24x + 5 = 0, we identify A = 3, B = �24 and C = 5. Here, we get:

x = �(�24) ±

p

(�24)2 � 4(3)(5) 2(3)

= 24 ±

p 516

6

Since p

516 = 2 p

129, this reduces to x = 12± p

129 3 .

3Think about what �(3 ± p

15) is really telling you.

0.7 Quadratic Equations 87

It is worth noting that the Quadratic Formula applies to all quadratic equations - even ones we could solve using other techniques. For example, to solve 2x2 + 5x �3 = 0 we identify A = 2, B = 5 and C = �3. This yields:

x = �5 ±

p

52 � 4(2)(�3) 2(2)

= �5 ±

p 49

4 = �5 ± 7

4

At this point, we have x = �5+74 = 1 2 and x =

�5�7 4 =

�12 4 = �3 - the same two answers we obtained

factoring. We can also use it to solve x2 = 3, if we wanted to. From x2 � 3 = 0, we have A = 1, B = 0 and C = �3. The Quadratic Formula produces

x = �0 ±

p

02 � 4(1)(3) 2(1)

= ± p

12 2

= ±2 p

3 2

= ± p

3

As this last example illustrates, while the Quadratic Formula can be used to solve every quadratic equation, that doesn’t mean it should be used. Many times other methods are more efficient. We now provide a more comprehensive approach to solving Quadratic Equations.

Strategies for Solving Quadratic Equations

• If the variable appears in the squared term only, isolate it and Extract Square Roots.

• Otherwise, put the nonzero terms on one side of the equation so that the other side is 0.

– Try factoring. – If the expression doesn’t factor easily, use the Quadratic Formula.

The reader is encouraged to pause for a moment to think about why ‘Completing the Square’ doesn’t appear in our list of strategies despite the fact that we’ve spent the majority of the section so far talking about it.4 Let’s get some practice solving quadratic equations, shall we?

Example 0.7.1. Find all real number solutions to the following equations.

1. 3 � (2w � 1)2 = 0 2. 5x � x(x � 3) = 7 3. (y � 1)2 = 2 � y + 2 3

4. 5(25 � 21x) = 59 4

� 25x2 5. �4.9t2 + 10t p

3 + 2 = 0 6. 2x2 = 3x4 � 6

Solution.

1. Since 3� (2w �1)2 = 0 contains a perfect square, we isolate it first then extract square roots: 3 � (2w � 1)2 = 0

3 = (2w � 1)2 Add (2w � 1)2 ± p

3 = 2w � 1 Extract Square Roots 1 ±

p 3 = 2w Add 1

1 ± p

3 2

= w Divide by 2

4Unacceptable answers include “Jeff and Carl are mean” and “It was one of Carl’s Pedantic Rants”.

88 Prerequisites

We find our two answers w = 1± p

3 2 . The reader is encouraged to check both answers by

substituting each into the original equation.5

2. To solve 5x � x(x � 3) = 7, we begin performing the indicated operations and getting one side equal to 0.

5x � x(x � 3) = 7 5x � x2 + 3x = 7 Distribute

�x2 + 8x = 7 Gather like terms �x2 + 8x � 7 = 0 Subtract 7

At this point, we attempt to factor and find �x2 + 8x � 7 = (x � 1)(�x + 7). Using the Zero Product Property, we get x � 1 = 0 or �x + 7 = 0. Our answers are x = 1 or x = 7, both of which are easy to check.

3. Even though we have a perfect square in (y � 1)2 = 2 � y+23 , Extracting Square Roots won’t help matters since we have a y on the other side of the equation. Our strategy here is to perform the indicated operations (and clear the fraction for good measure) and get 0 on one side of the equation.

(y � 1)2 = 2 � y + 2 3

y2 � 2y + 1 = 2 � y + 2 3

Perfect Square Trinomial

3(y2 � 2y + 1) = 3 ✓

2 � y + 2 3

◆

Multiply by 3

3y2 � 6y + 3 = 6 � 3 ✓

y + 2 3

◆

Distribute

3y2 � 6y + 3 = 6 � (y + 2) 3y2 � 6y + 3 � 6 + (y + 2) = 0 Subtract 6, Add (y + 2)

3y2 � 5y � 1 = 0

A cursory attempt at factoring bears no fruit, so we run this through the Quadratic Formula with A = 3, B = �5 and C = �1.

y = �(�5) ±

p

(�5)2 � 4(3)(�1) 2(3)

y = 5 ±

p 25 + 12 6

y = 5 ±

p 37

6

Since 37 is prime, we have no way to reduce p

37. Thus, our final answers are y = 5± p

37 6 .

The reader is encouraged to supply the details of the challenging verification of the answers. 5It’s excellent practice working with radicals fractions so we really, really want you to take the time to do it.

0.7 Quadratic Equations 89

4. We proceed as before; our aim is to gather the nonzero terms on one side of the equation.

5(25 � 21x) = 59 4

� 25x2

125 � 105x = 59 4

� 25x2 Distribute

4(125 � 105x) = 4 ✓

59 4

� 25x2 ◆

Multiply by 4

500 � 420x = 59 � 100x2 Distribute

500 � 420x � 59 + 100x2 = 0 Subtract 59, Add 100x2

100x2 � 420x + 441 = 0 Gather like terms

With highly composite numbers like 100 and 441, factoring seems inefficient at best,6 so we apply the Quadratic Formula with A = 100, B = �420 and C = 441:

x = �(�420) ±

p

(�420)2 � 4(100)(441) 2(100)

= 420 ±

p 176000 � 176400

200

= 420 ±

p 0

200

= 420 ± 0

200

= 420 200

= 21 10

To our surprise and delight we obtain just one answer, x = 2110 .

5. Our next equation �4.9t2 + 10t p

3 + 2 = 0, already has 0 on one side of the equation, but with coefficients like �4.9 and 10

p 3, factoring with integers is not an option. We could make

things a bit easier on the eyes by clearing the decimal (by multiplying through by 10) to get �49t2 +100t

p 3+20 = 0 but we simply cannot rid ourselves of the irrational number

p 3. The

Quadratic Formula is our only recourse. With A = �49, B = 100 p

3 and C = 20 we get:

6This is actually the Perfect Square Trinomial (10x � 21)2.

90 Prerequisites

t = �100

p 3 ±

q

(100 p

3)2 � 4(�49)(20) 2(�49)

= �100

p 3 ±

p 30000 + 3920

�98

= �100

p 3 ±

p 33920

�98

= �100

p 3 ± 8

p 530

�98

= 2(�50

p 3 ± 4

p 530)

2(�49)

= �50

p 3 ± 4

p 530

�49 Reduce

= �(�50

p 3 ± 4

p 530)

49 Properties of Negatives

= 50

p 3 ⌥ 4

p 530

49 Distribute

You’ll note that when we ‘distributed’ the negative in the last step, we changed the ‘±’ to a ‘⌥.’ While this is technically correct, at the end of the day both symbols mean ‘plus or minus’,7

so we can write our answers as t = 50 p

3±4 p

530 49 . Checking these answers are a true test of

arithmetic mettle.

6. At first glance, the equation 2x2 = 3x4 � 6 seems misplaced. The highest power of the variable x here is 4, not 2, so this equation isn’t a quadratic equation - at least not in terms of the variable x . It is, however, an example of an equation that is quadratic ‘in disguise.’8 We introduce a new variable u to help us see the pattern - specifically we let u = x2. Thus u2 = (x2)2 = x4. So in terms of the variable u, the equation 2x2 = 3x4 � 6 is 2u = 3u2 � 6. The latter is a quadratic equation, which we can solve using the usual techniques:

2u = 3u2 � 6 0 = 3u2 � 2u � 6 Subtract 2u

After a few attempts at factoring, we resort to the Quadratic Formula with A = 3, B = �2,

7There are instances where we need both symbols, however. For example, the Sum and Difference of Cubes Formulas (page 71) can be written as a single formula: a3 ± b3 = (a ± b)(a2 ⌥ ab + b2). In this case, all of the ‘top’ symbols are read to give the sum formula; the ‘bottom’ symbols give the difference formula.

8More formally, quadratic in form. Carl likes ‘Quadratics in Disguise’ since it reminds him of the tagline of one of his beloved childhood cartoons and toy lines.

0.7 Quadratic Equations 91

C = �6 and get:

u = �(�2) ±

p

(�2)2 � 4(3)(�6) 2(3)

= 2 ±

p 4 + 72 6

= 2 ±

p 76

6

= 2 ±

p 4 · 19

6

= 2 ± 2

p 19

6 Properties of Radicals

= 2(1 ±

p 19)

2(3) Factor

= 1 ±

p 19

3 Reduce

We’ve solved the equation for u, but what we still need to solve the original equation9 - which means we need to find the corresponding values of x . Since u = x2, we have two equations:

x2 = 1 +

p 19

3 or x2 =

1 � p

19 3

We can solve the first equation by extracting square roots to get x = ± q

1+ p

19 3 . The second

equation, however, has no real number solutions because 1� p

19 3 is a negative number. For

our final answers we can rationalize the denominator10 to get:

x = ±

s

1 + p

19 3

= ±

s

1 + p

19 3

· 3 3

= ± p

3 + 3 p

19 3

As with the previous exercise, the very challenging check is left to the reader.

Our last example above, the ‘Quadratic in Disguise’, hints that the Quadratic Formula is applicable to a wider class of equations than those which are strictly quadratic. We give some general guidelines to recognizing these beasts in the wild on the next page.

9Or, you’ve solved the equation for ‘you’ (u), now you have to solve it for your instructor (x). 10We’ll say more about this technique in Section 0.9.

92 Prerequisites

Identifying Quadratics in Disguise An equation is a ‘Quadratic in Disguise’ if it can be written in the form: AX 2m + BX m + C = 0. In other words:

• There are exactly three terms, two with variables and one constant term.

• The exponent on the variable in one term is exactly twice the variable on the other term.

To transform a Quadratic in Disguise to a quadratic equation, let u = X m so u2 = (X m)2 = X 2m. This transforms the equation into Au2 + Bu + C = 0.

For example, 3x6 � 2x3 + 1 = 0 is a Quadratic in Disguise, since 6 = 2 · 3. If we let u = x3, we get u2 = (x3)2 = x6, so the equation becomes 3u2 � 2u + 1 = 0. However, 3x6 � 2x2 + 1 = 0 is not a Quadratic in Disguise, since 6 6= 2 · 2. The substitution u = x2 yields u2 = (x2)2 = x4, not x6 as required. We’ll see more instances of ‘Quadratics in Disguise’ in later sections.

We close this section with a review of the discriminant of a quadratic equation as defined below.

Definition 0.16. The Discriminant: Given a quadratic equation AX 2 + BX + C = 0, the quantity B2 � 4AC is called the discriminant of the equation.

The discriminant is the radicand of the square root in the quadratic formula:

X = �B ±

p B2 � 4AC 2A

It discriminates between the nature and number of solutions we get from a quadratic equation. The results are summarized below.

Theorem 0.10. Discriminant Theorem: Given a Quadratic Equation AX 2 + BX + C = 0, let D = B2 � 4AC be the discriminant.

• If D > 0, there are two distinct real number solutions to the equation.

• If D = 0, there is one repeated real number solution.

Note: ‘Repeated’ here comes from the fact that ‘both’ solutions �B±02A reduce to � B 2A .

• If D < 0, there are no real solutions.

For example, x2 + x � 1 = 0 has two real number solutions since the discriminant works out to be (1)2 � 4(1)(�1) = 5 > 0. This results in a ±

p 5 in the Quadratic Formula, generating two different

answers. On the other hand, x2 + x + 1 = 0 has no real solutions since here, the discriminant is (1)2 � 4(1)(1) = �3 < 0 which generates a ±

p �3 in the Quadratic Formula. The equation

x2 + 2x + 1 = 0 has discriminant (2)2 � 4(1)(1) = 0 so in the Quadratic Formula we get a ± p

0 = 0 thereby generating just one solution. More can be said as well. For example, the discriminant of 6x2 � x � 40 = 0 is 961. This is a perfect square,

p 961 = 31, which means our solutions are

0.7 Quadratic Equations 93

rational numbers. When our solutions are rational numbers, the quadratic actually factors nicely. In our example 6x2 � x � 40 = (2x + 5)(3x � 8). Admittedly, if you’ve already computed the discriminant, you’re most of the way done with the problem and probably wouldn’t take the time to experiment with factoring the quadratic at this point – but we’ll see another use for this analysis of the discriminant in the next section.11

11Specifically in Example 0.8.1.

94 Prerequisites

0.7.1 Exercises

In Exercises 1 - 21, find all real solutions. Check your answers, as directed by your instructor.

1. 3 ✓

x � 1 2

◆2 =

5 12

2. 4 � (5t + 3)2 = 3 3. 3(y2 � 3)2 � 2 = 10

4. x2 + x � 1 = 0 5. 3w2 = 2 � w 6. y (y + 4) = 1

7. z 2

= 4z2 � 1 8. 0.1v2 + 0.2v = 0.3 9. x2 = x � 1

10. 3 � t = 2(t + 1)2 11. (x � 3)2 = x2 + 9 12. (3y � 1)(2y + 1) = 5y

13. w4 + 3w2 � 1 = 0 14. 2x4 + x2 = 3 15. (2 � y )4 = 3(2 � y )2 + 1

16. 3x4 + 6x2 = 15x3 17. 6p + 2 = p2 + 3p3 18. 10v = 7v3 � v5

19. y2 � p

8y = p

18y � 1 20. x2 p

3 = x p

6 + p

12 21. v2

3 =

v p

3 2

+ 1

In Exercises 22 - 27, find all real solutions and use a calculator to approximate your answers, rounded to two decimal places.

22. 5.542 + b2 = 36 23. ⇡r2 = 37 24. 54 = 8r p

2 + ⇡r2

25. �4.9t2 + 100t = 410 26. x2 = 1.65(3 � x)2 27. (0.5+2A)2 = 0.7(0.1�A)2

In Exercises 28 - 30, use Theorem 0.3 along with the techniques in this section to find all real solutions to the following.

28. |x2 � 3x | = 2 29. |2x � x2| = |2x � 1| 30. |x2 � x + 3| = |4 � x2|

31. Prove that for every nonzero number p, x2 + xp + p2 = 0 has no real solutions.

32. Solve for t : �1 2

gt2 + vt + h = 0. Assume g > 0, v � 0 and h � 0.

0.7 Quadratic Equations 95

0.7.2 Answers

1. x = 3 ±

p 5

6 2. t = �4

5 ,�2

5 3. y = ±1, ±

p 5

4. x = �1 ±

p 5

2 5. w = �1, 2

3 6. y = �2 ±

p 5

7. z = 1 ±

p 65

16 8. v = �3, 1 9. No real solution.

10. t = �5 ±

p 33

4 11. x = 0 12. y =

2 ± p

10 6

13. w = ± rp

13 � 3 2

14. x = ±1 15. y = 4 ± p

6 + 2 p

13 2

16. x = 0, 5 ±

p 17

2 17. p = �1

3 ,±

p 2 18. v = 0,±

p 2,±

p 5

19. y = 5 p

2 ± p

46 2

20. x = p

2 ± p

10 2

21. v = � p

3 2

, 2 p

3

22. b = ± p

13271 50

⇡ ±2.30 23. r = ± r

37 ⇡

⇡ ±3.43

24. r = �4

p 2 ±

p 54⇡ + 32

⇡ , r ⇡ �6.32, 2.72 25. t = 500 ± 10

p 491

49 , t ⇡ 5.68, 14.73

26. x = 99 ± 6

p 165

13 , x ⇡ 1.69, 13.54 27. A = �107 ± 7

p 70

330 , A ⇡ �0.50,�0.15

28. x = 1, 2, 3 ±

p 17

2 29. x = ±1, 2 ±

p 3 30. x = �1

2 , 1, 7

31. The discriminant is: D = p2 � 4p2 = �3p2 < 0. Since D < 0, there are no real solutions.

32. t = v ±

p

v2 + 2gh g