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Quadratic Equations: Factoring and Solving

Learn how to factor and solve quadratic equations with step-by-step examples and special patterns. Find zeros, factor trinomials, and solve real-world problems using quadratic equations. Increase your understanding and mastery of quadratic equations.

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Quadratic Equations: Factoring and Solving

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  1. ANSWER Notice thatm = – 4andn = – 5. So, x2–9x + 20 = (x –4)(x –5). EXAMPLE 1 Factor trinomials of the form x2+ bx + c Factor the expression. a. x2 – 9x + 20 b. x2 + 3x – 12 SOLUTION a.You wantx2 – 9x + 20 = (x + m) (x + n)where mn= 20 andm + n = –9.

  2. ANSWER Notice that there are no factors mand nsuch that m + n = 3. So, x2 + 3x – 12 cannot be factored. EXAMPLE 1 Factor trinomials of the form x2+ bx + c b.You wantx2 + 3x – 12 = (x + m) (x + n)where mn= – 12 andm + n = 3.

  3. EXAMPLE 2 Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of two squares = (x + 7) (x –7) b. d 2 + 12d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 Perfect square trinomial = z2 – 2(z) (13) + 132 = (z –13)2

  4. x – 9 = 0 or x + 4 = 0 x = 9 or x = – 4 ANSWER The correct answer is C. EXAMPLE 3 Standardized Test Practice SOLUTION x2 – 5x – 36 = 0 Write original equation. (x – 9)(x + 4) = 0 Factor. Zero product property Solve for x.

  5. 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. EXAMPLE 4 Use a quadratic equation as a model Nature Preserve

  6. or x –200 = 0 x + 1200 = 0 or x = 200 x = –1200 EXAMPLE 4 Use a quadratic equation as a model 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. Zero product property Solve for x.

  7. EXAMPLE 4 Use a quadratic equation as a model ANSWER 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.

  8. EXAMPLE 5 Find the zeros of quadratic functions. Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 SOLUTION a. y = x2 – x – 12 Write original function. = (x + 3) (x – 4) Factor. The zeros of the function are –3 and 4. CheckGraph y = x2 – x – 12. The graph passes through (–3, 0) and (4, 0).

  9. EXAMPLE 5 Find the zeros of quadratic functions. Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 SOLUTION b. y = x2 + 12x + 36 Write original function. = (x + 6) (x + 6) Factor. The zeros of the function is – 6 CheckGraph y = x2 + 12x + 36. The graph passes through ( – 6, 0).

  10. EXAMPLE 1 Factor ax2 + bx + c where c > 0 Factor 5x2 – 17x + 6. SOLUTION You want 5x2 – 17x + 6 = (kx + m) (lx + n) where kand lare factors of 5 andmand nare factors of 6. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign. So, mand nmust both be negative because the coefficient of x, – 17, is negative.

  11. ANSWER The correct factorization is5x2 – 17x + 6 = (5x – 2) (x – 3). EXAMPLE 1 Factor ax2 + bx + c where c > 0

  12. ANSWER The correct factorization is3x2 + 20x – 7= (3x – 1) (x + 7). EXAMPLE 2 Factor ax2 +bx + c where c < 0 Factor 3x2 + 20x – 7. SOLUTION You want3x2 + 20x – 7 = (kx + m) (lx + n)wherekandlarefactors of3andmandnare factors of– 7. Becausemn < 0, mandn have opposite signs.

  13. EXAMPLE 3 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. 36w2 – 12w + 1 = (6w)2 – 2(6w) (1) + (1)2 Perfect square trinomial = (6w – 1)2

  14. EXAMPLE 4 Factor out monomials first Factor the expression. = 5(x2 – 9) a. 5x2 – 45 = 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)

  15. orx + 4 = 0 3x – 2 = 0 x = orx = – 4 23 EXAMPLE 5 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. Zero product property Solve for x.

  16. EXAMPLE 5 Solve quadratic equations (b) 5p2 – 16p + 15 = 4p – 5. 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.

  17. You have made a rectangular quilt that is 5 feet by 4 feet. You want to use the remaining 10square feet of fabric to add a decorative border of uniform width to the quilt. What should the width of the quilt’s border be? EXAMPLE 6 Use a quadratic equation as a model Quilts

  18. orx + 5 = 0 2x – 1 = 0 12 x = orx = – 5 EXAMPLE 6 Use a quadratic equation as a model SOLUTION Write a verbal model. Then write an equation. 10 = 20 + 18x + 4x2 – 20 Multiply using FOIL. 0 = 4x2 + 18x – 10 Write in standard form 0 = 2x2 + 9x – 5 Divide each side by 2. 0 = (2x – 1) (x + 5) Factor. Zero product property Solve for x.

  19. ANSWER Reject the negative value,– 5. The border’s width should beft, or6 in. 12 EXAMPLE 6 Use a quadratic equation as a model

  20. A monthly teen magazine has 28,000 subscribers when it charges $10 per annual subscription. For each $1 increase in price, the magazine loses about 2000 subscribers. How much should the magazine charge to maximize annual revenue ? What is the maximum annual revenue ? EXAMPLE 7 Solve a multi-step problem Magazines

  21. EXAMPLE 7 Solve a multi-step problem SOLUTION STEP 1 Define the variables. Let xrepresent the price increase and R(x) represent the annual revenue. STEP 2 Write a verbal model. Then write and simplify a quadratic function.

  22. EXAMPLE 7 Solve a multi-step problem R(x) = (– 2000x + 28,000) (x + 10) R(x) = – 2000(x – 14) (x + 10)

  23. 14 + (– 1 0) 2 EXAMPLE 7 Solve a multi-step problem Identify the zeros and find their average. Find how much each subscription should cost to maximize annual revenue. STEP 3 The zeros of the revenue function are 14 and –10. The average of the zeroes is =2. To maximize revenue, each subscription should cost $10 + $2= $12. Find the maximum annual revenue. STEP 4 R(2) = – 2000(2– 14) (2+ 10) = $288,000

  24. ANSWER The magazine should charge $12 per subscription to maximize annual revenue. The maximum annual revenue is $288,000. EXAMPLE 7 Solve a multi-step problem

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