# 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 PresentationTRANSCRIPT

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,)