HW: Pg. 267 #47-67o, 69, 70. 5.2 Solving Quadratic Equations by Finding Square Roots.
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Transcript of 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