1.3 solving quadratic equations by factoring

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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). - PowerPoint PPT Presentation

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

  • Vocabulary ReminderMonomial

    Binomial

    TrinomialAn 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)

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

  • Factor the expression.a. x2 9x + 20SOLUTIONa. You want x2 9x + 20 = (x + m) (x + n) where mn = 20 and m + n = 9.

  • b. x2 + 3x 12

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

  • Guided PracticeFactor the expression. If the expression cannot be factored, say so.1. x2 3x 18SOLUTIONYou want x2 3x 18 = (x + m) (x + n) where mn = 18 and m + n = 3.

    Factor of 18 : m, n 1, 18 1, 18 3, 6 2, 92, 9 6, 3Sum of factors: m + n 171737 7 3

  • Guided Practice2. n2 3n + 9SOLUTIONYou want n2 3n + 9 = (x + m) (x + n) where mn = 9 and m + n = 3.

    Factor of 9 : m, n 1, 9 1, 93, 3 3, 3Sum of factors: m + n10106 6

  • Factor with Special PatternsFactor the expression.a. x2 49= (x + 7) (x 7)Difference of two squaresb. d 2 + 12d + 36= (d + 6)2Perfect square trinomialc. z2 26z + 169= (z 13)2Perfect square trinomial= x2 72= d 2 + 2(d)(6) + 62= z2 2(z) (13) + 132

  • Guided Practice4. x2 9= (x 3) (x + 3)Difference of two squares5. q2 100= (q 10) (q + 10)Difference of two squares6. y2 + 16y + 64= (y + 8)2Perfect square trinomialFactor the expression.= x2 32= q2 102= y2 + 2(y) 8 + 82

  • Zero Product PropertyLet 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 roots or zeros of the equation.

  • Use a Quadratic Equation as a ModelNature Preserve

  • SOLUTION480,000 = 240,000 + 1000x + x2Multiply using FOIL.0 = x2 + 1000x 240,000Write in standard form.0 = (x 200) (x + 1200)Factor.Zero product propertySolve for x.ANSWERReject the negative value, 1200. The fields length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

  • What If ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field.SOLUTION600000 =Multiply using FOIL.300000 + 1000x + 300x + x20 = x2 + 1300x 300000Write in standard form.0 = (x 200) (x + 1500)Factor.Zero product propertyx = 200orx = 1500Solve for x.

  • Finding the Zeros of an EquationThe 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.

  • Example: Find the Zeros ofy=x2-x-6y=x2-x-6Change y to 00=x2-x-6Factor the right side0=(x-3)(x+2)Set factors =0x-3=0 OR x+2=0Solve each equation +3 +3 -2 -2 x=3 OR x=-2Check your solutions!

    If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).

  • Guided PracticeFind the zeros of the function by rewriting the function in intercept form. y = x2 + 5x 14SOLUTION y = x2 + 5x 14Write original function.= (x + 7) (x 2)Factor.The zeros of the function is 7 and 2Check Graph y = x2 + 5x 14. The graph passes through ( 7, 0) and (2, 0).

  • Assignmentp. 21, 3-54 every 3rd problem(3,6,9,12,)

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