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1.3 Solving Quadratic Equations by Factoring

1.3 Solving Quadratic Equations by Factoring. (p. 18) How can factoring be used to solve quadratic equation when a=1?. Vocabulary Reminder. Monomial Binomial Trinomial. An expression that is either a number, a variable, or the product of a number and one or more variables. (4, x, 5y).

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1.3 Solving Quadratic Equations by Factoring

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  1. 1.3 Solving Quadratic Equations by Factoring (p. 18) How can factoring be used to solve quadratic equation when a=1?

  2. Vocabulary Reminder • Monomial • Binomial • Trinomial An expression that is either a number, a variable, or the product of a number and one or more variables. (4, x, 5y) The sum of two monomials (x+4)

  3. To solve a quadratic eqn. by factoring, you must remember your factoring patterns!

  4. ANSWER Notice thatm = – 4andn = – 5. So, x2–9x + 20 = (x –4)(x –5). Factor the expression. a. x2 – 9x + 20 SOLUTION a.You wantx2 – 9x + 20 = (x + m) (x + n)where mn= 20 andm + n = –9.

  5. ANSWER Notice that there are no factors mand nsuch that m + n = 3. So, x2 + 3x – 12 cannot be factored. Factor b.You wantx2 + 3x – 12 = (x + m) (x + n)where mn= – 12 andm + n = 3. b. x2 + 3x – 12

  6. ANSWER m = – 6 and n = 3 so x2 – 3x – 18 = (x – 6) (x + 3) Guided Practice Factor the expression. If the expression cannot be factored, say so. 1. x2 – 3x – 18 SOLUTION You wantx2 – 3x – 18 = (x + m) (x + n)where mn= – 18andm + n = –3.

  7. ANSWER Notice that there are no factors mand nsuch that m + n = – 3 . So, n2 – 3x + 9 cannot be factored. Guided Practice 2. n2 – 3n + 9 SOLUTION You wantn2 – 3n + 9 = (x + m) (x + n)where mn= 9 andm + n = –3.

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

  9. Guided Practice Factor the expression. 4. x2 – 9 = x2 – 32 Difference of two squares = (x –3) (x +3) Difference of two squares 5. q2 – 100 = q2 – 102 = (q – 10) (q + 10) 6. y2 + 16y + 64 = y2 + 2(y) 8 + 82 = (y +8)2 Perfect square trinomial

  10. Zero Product Property • Let A and B be real numbers or algebraic expressions. If AB=0, then A=0 or B=0. • This means that If the product of 2 factors is zero, then at least one of the 2 factors had to be zero itself! • The answers, the solutions of a quadratic equation are called the rootsorzeros of the equation.

  11. 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. Use a Quadratic Equation as a Model Nature Preserve

  12. or x –200 = 0 x + 1200 = 0 or x = 200 x = –1200 SOLUTION Multiply using FOIL. 480,000 = 240,000 + 1000x + x2 Write in standard form. 0 = x2+ 1000x –240,000 Factor. 0 = (x –200) (x + 1200) Zero product property Solve for x. 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.

  13. New width (meters) New Area New Length (meters) = 2(1000)(300) (1000 + x) = (300 + x) or x –200 = 0 x + 1500 = 0 What If ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field. SOLUTION 600000 = 300000 + 1000x + 300x + x2 Multiply using FOIL. 0 = x2+ 1300x –300000 Write in standard form. 0 = (x –200) (x + 1500) Factor. Zero product property or Solve for x. x = –1500 x = 200

  14. ANSWER Reject the negative value, – 1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 1200 meters by 500 meters.

  15. Finding the Zeros of an Equation • The Zeros of an equation are the x-intercepts ! • First, change y to a zero. • Rewrite the function in intercept form. • Now, solve for x. • The solutions will be the zeros of the equation.

  16. Example: Find the Zeros ofy=x2-x-6 y=x2-x-6 Change y to 0 0=x2-x-6 Factor the right side 0=(x-3)(x+2) Set factors =0 x-3=0 OR x+2=0 Solve each equation +3 +3 -2 -2 x=3 OR x=-2 Check your solutions! If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).

  17. Guided Practice Find the zeros of the function by rewriting the function in intercept form. y = x2 + 5x – 14 SOLUTION y = x2 + 5x – 14 Write original function. = (x + 7) (x – 2) Factor. The zeros of the function is – 7 and 2 CheckGraph y = x2 + 5x – 14. The graph passes through ( – 7, 0) and (2, 0).

  18. Assignment p. 21, 3-54 every 3rd problem (3,6,9,12,…)

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