HW: Pg. 267 #47-67o, 69, 70

5.2 Solving Quadratic Equations by Finding Square Roots

EXAMPLE 1 Factor trinomials of the form x2 + bx + c

Factor the expression.

a. x2 – 9x + 20

b. x2 + 3x – 12

SOLUTION

a. You want x2 – 9x + 20 = (x + m)(x + n) where mn = 20 and m + n = –9.

ANSWER

Notice that m = –4 and n = –5. So, x2 – 9x + 20 = (x – 4)(x – 5).

EXAMPLE 1 Factor trinomials of the form x2 + bx + c

b. You want x2 + 3x – 12 = (x + m)(x + n) where mn = – 12 and m + n = 3.

ANSWER

Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.

GUIDED PRACTICE for Example 1

Factor the expression. If the expression cannot be factored, say so.

1. x2 – 3x – 18

ANSWER

(x – 6)(x + 3)

2. n2 – 3n + 9

cannot be factored

ANSWER

3. r2 + 2r – 63

(r + 9)(r –7)

ANSWER

Special Factoring Patterns

PATTERN NAME PATTERN EXAMPLEDifference of Two Squares

(FL pattern)

Perfect Square Trinomial

(smiley face pattern)

2 2 ( )( )a b a b a b 2 9 ( )3)(3x x x

2 2 2

2 2 2

2 ( )

2 ( )

a ab b a b

a ab b a b

2 2

2 2

12 (36 6)

8 (6 4)1

x ab x

x ab x

EXAMPLE 2 Factor with special patterns

Factor the expression.

a. x2 – 49

= (x + 7)(x – 7)

Difference of two squares

b. d 2 + 12d + 36

= (d + 6)2

Perfect square trinomial

c. z2 – 26z + 169

= (z – 13)2

Perfect square trinomial

= x2 – 72

= d 2 + 2(d)(6) + 62

= z2 – 2(z) (13) + 132

GUIDED PRACTICE for Example 2

4. x2 – 9

(x – 3)(x + 3)

5. q2 – 100

(q – 10)(q + 10)

6. y2 + 16y + 64

(y + 8)2

Factor the expression.

ANSWER

ANSWER

ANSWER

7. w2 – 18w + 81

(w – 9)2ANSWER

EXAMPLE 3 Use a quadratic equation as a model

Nature Preserve

A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Find the new dimensions of the field.

SOLUTION

480,000 = 240,000 + 1000x + x2 Multiply using FOIL.0 = x2 + 1000x – 240,000 Write in standard form.0 = (x – 200) (x + 1200) Factor.

x – 200 = 0 x + 1200 = 0or Zero product propertyx = 200 or x = –1200 Solve for x.

Reject the negative value, –1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

EXAMPLE 4 Factor ax2 + bx + c where c > 0

Factor 5x2 – 17x + 6.

SOLUTION

You want 5x2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and m and n are factors of 6. You can assume that k and l are positive and k ≥ l. Because mn > 0, m and n have the same sign. So, m and n must both be negative because the coefficient of x, –17, is negative.

ANSWER

The correct factorization is 5x2 –17x + 6 = (5x – 2)(x – 3).

EXAMPLE 5 Factor ax2 + bx + c where c < 0

Factor 3x2 + 20x – 7.

SOLUTION

You want 3x2 + 20x – 7 = (kx + m)(lx + n) where k and l are factors of 3 and m and n are factors of –7. Because mn < 0, m and n have opposite signs.

ANSWER

The correct factorization is 3x2 + 20x – 7 = (3x – 1)(x + 7).

GUIDED PRACTICE for Examples 4 and 5GUIDED PRACTICE

Factor the expression. If the expression cannot be factored, say so.

8. 7x2 – 20x – 3

(7x + 1)(x – 3)

9. 5z2 + 16z + 3

10. 2w2 + w + 3

cannot be factored

ANSWER (5z + 1)(z + 3).

ANSWER

ANSWER

12. 4u2 + 12u + 5

(2u + 1)(2u + 5)

13. 4x2 – 9x + 2

(4x – 1)(x – 2)

11. 3x2 + 5x – 12

(3x – 4)(x + 3)ANSWER

ANSWER

ANSWER

EXAMPLE 6 Factor with special patterns

Factor the expression.

a. 9x2 – 64

= (3x + 8)(3x – 8)

Difference of two squares

b. 4y2 + 20y + 25

= (2y + 5)2

Perfect square trinomial

c. 36w2 – 12w + 1

= (6w – 1)2

Perfect square trinomial

= (3x)2 – 82

= (2y)2 + 2(2y)(5) + 52

= (6w)2 – 2(6w)(1) + (1)2

GUIDED PRACTICEGUIDED PRACTICE for Example 6

Factor the expression.

14. 16x2 – 1

(4x + 1)(4x – 1)

15. 9y2 + 12y + 4

(3y + 2)2

16. 4r2 – 28r + 49

(2r – 7)2

17. 25s2 – 80s + 64

(5s – 8)2ANSWER

ANSWER

ANSWER

ANSWER

18. 49z2 + 4z + 9

(7z + 3)2

19. 36n2 – 9 = (3y)2

(6n – 3)(6n +3)

ANSWER

ANSWER

EXAMPLE 7 Factor out monomials first

Factor the expression.

a. 5x2 – 45

= 5(x + 3)(x – 3)

b. 6q2 – 14q + 8

= 2(3q – 4)(q – 1)

c. –5z2 + 20z

d. 12p2 – 21p + 3

= 5(x2 – 9)

= 2(3q2 – 7q + 4)

= –5z(z – 4)

= 3(4p2 – 7p + 1)

GUIDED PRACTICEGUIDED PRACTICE for Example 7

Factor the expression.

20. 3s2 – 24

21. 8t2 + 38t – 10

2(4t – 1) (t + 5)

3(s2 – 8)

22. 6x2 + 24x + 15

3(2x2 + 8x + 5)

23. 12x2 – 28x – 24

4(3x + 2)(x – 3)

24. –16n2 + 12n

–4n(4n – 3)ANSWER

ANSWER

ANSWER

ANSWER

ANSWER

25. 6z2 + 33z + 36

3(2z + 3)(z + 4)ANSWER

EXAMPLE 8 Solve quadratic equations

Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5.

a. 3x2 + 10x – 8 = 0

(3x – 2)(x + 4) = 0

3x – 2 = 0 or x + 4 = 0

Write original equation.

Factor.

Zero product property

Solve for x.or x = –4x = 23

b. 5p2 – 16p + 15 = 4p – 5. Write original equation.5p2 – 20p + 20 = 0

p2 – 4p + 4 = 0

(p – 2)2 = 0

p – 2 = 0

p = 2

Write in standard form.

Divide each side by 5.

Factor.

Zero product property

Solve for p.

GUIDED PRACTICEGUIDED PRACTICE for Example 8

Solve the equation.

26. 6x2 – 3x – 63 = 0

or –3 3 12

27. 12x2 + 7x + 2 = x +8

no solution

28. 7x2 + 70x + 175 = 0

–5

ANSWER

ANSWER

ANSWER

Homework:

Pg. 260 #23-91 eoo

and

prepare for quiz on 5.3 & 5.2 next class