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

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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 2. n2 – 3n + 9 3. r2 + 2r – 63 ANSWER ANSWER ANSWER (r + 9)(r –7) (x – 6)(x + 3) cannot be factored

Special Factoring Patterns
PATTERN NAME PATTERN EXAMPLE Difference of Two Squares (FL pattern) Perfect Square Trinomial (smiley face pattern)

Difference of two squares
EXAMPLE 2 Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of two squares = (x + 7)(x – 7) b. d d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 = z2 – 2(z) (13) + 132 Perfect square trinomial = (z – 13)2

GUIDED PRACTICE for Example 2 Factor the expression. 4. x2 – 9 7. w2 – 18w + 81 ANSWER (x – 3)(x + 3) ANSWER (w – 9)2 5. q2 – 100 ANSWER (q – 10)(q + 10) 6. y2 + 16y + 64 ANSWER (y + 8)2

Multiply using FOIL. Write in standard form. Factor.
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, x + x2 Multiply using FOIL. 0 = x x – 240,000 Write in standard form. 0 = (x – 200) (x ) Factor. x – 200 = 0 x = 0 or Zero product property x = 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 GUIDED PRACTICE for Examples 4 and 5 Factor the expression. If the expression cannot be factored, say so. x2 + 5x – 12 x2 – 20x – 3 ANSWER (7x + 1)(x – 3) ANSWER (3x – 4)(x + 3) u2 + 12u + 5 z2 + 16z + 3 ANSWER (5z + 1)(z + 3). ANSWER (2u + 1)(2u + 5) w2 + w + 3 x2 – 9x + 2 ANSWER cannot be factored ANSWER (4x – 1)(x – 2)

Difference of two squares
EXAMPLE 6 Factor with special patterns Factor the expression. a. 9x2 – 64 = (3x)2 – 82 Difference of two squares = (3x + 8)(3x – 8) b. 4y2 + 20y + 25 = (2y)2 + 2(2y)(5) + 52 Perfect square trinomial = (2y + 5)2 c w2 – 12w + 1 = (6w)2 – 2(6w)(1) + (1)2 Perfect square trinomial = (6w – 1)2

GUIDED PRACTICE GUIDED PRACTICE for Example 6 Factor the expression. x2 – 1 z2 + 4z + 9 ANSWER (4x + 1)(4x – 1) ANSWER (7z + 3)2 y2 + 12y + 4 n2 – 9 = (3y)2 ANSWER (3y + 2)2 ANSWER (6n – 3)(6n +3) r2 – 28r + 49 ANSWER (2r – 7)2 s2 – 80s + 64 ANSWER (5s – 8)2

EXAMPLE 7 Factor out monomials first Factor the expression. a. 5x2 – 45 = 5(x2 – 9) = 5(x + 3)(x – 3) b. 6q2 – 14q + 8 = 2(3q2 – 7q + 4) = 2(3q – 4)(q – 1) c. –5z2 + 20z = –5z(z – 4) d. 12p2 – 21p + 3 = 3(4p2 – 7p + 1)

GUIDED PRACTICE GUIDED PRACTICE for Example 7 Factor the expression. s2 – 24 z2 + 33z + 36 ANSWER 3(s2 – 8) ANSWER 3(2z + 3)(z + 4) t2 + 38t – 10 ANSWER 2(4t – 1) (t + 5) x2 + 24x + 15 ANSWER 3(2x2 + 8x + 5) x2 – 28x – 24 ANSWER 4(3x + 2)(x – 3) –16n2 + 12n ANSWER –4n(4n – 3)

Write original equation.
EXAMPLE 8 Solve quadratic equations Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5. a. 3x2 + 10x – 8 = 0 Write original equation. (3x – 2)(x + 4) = 0 Factor. 3x – 2 = 0 or x + 4 = 0 Zero product property or x = –4 x = 23 Solve for x. b. 5p2 – 16p + 15 = 4p – 5. Write original equation. 5p2 – 20p + 20 = 0 Write in standard form. p2 – 4p + 4 = 0 Divide each side by 5. (p – 2)2 = 0 Factor. p – 2 = 0 Zero product property p = 2 Solve for p.

GUIDED PRACTICE GUIDED PRACTICE for Example 8 Solve the equation. x2 – 3x – 63 = 0 or –3 3 12 ANSWER x2 + 7x + 2 = x +8 ANSWER no solution x2 + 70x = 0 ANSWER –5

Pg. 260 #23-91 eoo and prepare for quiz on 5.3 & 5.2 next class
Homework: Pg. 260 #23-91 eoo and prepare for quiz on 5.3 & 5.2 next class