+Strategic

Intervention Material

Mathematics IX

Prepared by: Brian M. Mary T-I

Solving Quadratic Equation by Factoring

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Guide CardLEAST MASTERED SKILLS

Solving Quadratic Equation by Factoring

Sub Tasks Identifying quadratic equations Rewriting quadratic equations to its

standard form Factor trinomials in the form x2 + bx + c Determine roots of quadratic equation

ax2 + bx + c = 0, by factoring

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OverviewA quadratic equation in one variable is a mathematical sentence of degree 2 that can be written in the following form

ax2 + bx + c = 0,

where a, b, and c are real numbers and a ≠ 0.

Why do you think

‘a’ must not be

equal to zero in

the equation

ax2 + bx + c = 0?

How are quadratic equations used in solving real – life problems and in making decisions?

Many formulas used in the physical world are quadratic in nature since they become second-degree equations when solving for one of the variables. Likewise, many word problems require the use of the quadratic equation.

At the enrichment card, we will consider some of the common uses of the quadratic equations.

+ Activity # 1

__________ 1. 3m + 8 = 15

__________ 2. x2 – 5x – 10 = 0

__________ 3. 2t2 – 7t = 12

__________ 4. 12 – 4x = 0

__________ 5. 25 – r2 = 4r

Quadratic or Not Quadratic?Direction. Identify which of the following equations are quadratic and which are not. Write QE if the equations are quadratic and NQE if not quadratic equation.

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Activity # 2Set Me to Your Standard!

Direction. Write each quadratic equation in standard form, ax2 + bx + c = 0.

1. 3x – 2x2 = 7 ____________________ 2. 5 – 2r2 = 6r ____________________ 3. 2x(x – 3) = 15____________________ 4. (x + 3)(x + 4)= 0 ____________________ 5. (x + 4)2 + 8 = 0 ____________________

+ Activity # 3What Made Me?

We learned how to multiply two binomials as follows:

factors

(x+2)(x+6) = x2 + 6x + 2x + 12 = x2 + 8x + 12.

termsM u l t i p l y i n g

factorstermsF a c t o r i n g

x2 + 8x + 12 = (x + 2)(x + 6)

In factoring, we reverse the operation

The following will enable us to see how a trinomial factors.

x2 + 8x + 12 = (x + 2)(x + 6)

12 = 2 (6)

8 = 2 + 6

Product

Sum

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In general, the trinomial x2 + bx + c will factor only if there are two integers, which will we call m and n, such that m + n = b and m(n) = c.

Sum Productm + n m(n)

x2 + bx + c = (x + m)(x + n)

1. a2 + 11a + 18 m + n = 11m(n) = 18 2 + 9 = 11 2(9) = 18

The m and n values are 2 and 9. the factorization is,

a2 + 11a + 18 = (x + 2) (x + 9)

2. b2 – 2b – 15 m + n = - 2 m(n) = - 15

3 + (-5) = - 2 3(-5) = - 15

The m and n values are 3 and - 5. the factorization is,

b2 – 2b – 15 = (x + 3) (x – 5)

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Factor the following trinomial in the form x2 + bx + c.

x2 + bx + c m + n m(n) (x + m)(x + n)

x2 + 4x – 12 6 + (-2) 6(-2) (x + 6)(x – 2)

w2 – 8w + 12

x2 + 5x - 24

c2 + 6c + 5

r2 + 5r – 14

x2 + 5x + 7

After learning how to factor trinomial in the form x2 + bx + c,we will now determine roots of a quadratic equation using factoring.

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Activity # 4Factor then Solve!

Some quadratic equations can be solved easily by factoring. To solve each equations, the following procedures can be followed.

1. Transform the quadratic equation into standard form if necessary.2. Factor the quadratic expression.3. Set each factor of the quadratic expression equal to 0.4. Solve each resulting equation.

Example. Find the solution of x2 + 9x = -8 by factoring.

a. Transform the equation into standard formx2 + 9x = -8 x2 + 9x + 8 = 0

b. Factor the quadratic expression x2 + 9x + 8 = 0 (x + 1)(x +8) = 0

c. Set each factor equal to 0. (x + 1)(x + 8) = 0 x + 1 =

0; x + 8 = 0d. Solve each resulting equation.x + 1 = 0 x + 1 – 1

= 0 -1

x = - 1

x + 8 = 0 x + 8 – 8 = 0 - 8

x = - 8

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Direction. Determine the roots of the following quadratic equations using factoring.

1. x2 + 8x + 16 = 0 _______________________________________________________________________________________________________________

2. x2 – 9x – 14 = 0 _______________________________________________________________________________________________________________

3. y2 + 9y + 20 = 0 _______________________________________________________________________________________________________________

4. b2 – 10b + 21 = 0 _______________________________________________________________________________________________________________

Assessment

Card

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Enrichment Card

Mastery Points!Can you Determine two integers whose product is one number and whose sum is another number?Recognize when the trinomial x2 + bx + c will factor and when it will not?Factor trinomial of the form x2 + bx +c ?Determine roots of a quadratic function in the form ax2 + bx + c?

Number Theory* The product of two consecutive even numbers is 168. What are the integers?

Solution: Let x = the lesser even integer. Then, x + 2 = the next consecutive even integer.

Note: Consecutive even or odd integers are given by x, x+ 2, x + 4, …

product of two consecutive even integers is 168

x ( x + 2 )= 168

x2 + 2x = 168original equation

x2 + 2x – 168 = 0 write in standard form

( x + 14 ) ( x – 12 ) = 0 factor the left memberx + 14 = 0 or x – 12 = 0 set each factor equal to

zerox = - 14 x = 12 solve each equation

when x = - 14, then x + 2 = - 14 + 2 = - 12 and when x = 12, then x + 2 = 12 + 2 = 14

since (-14)(-12) = 168 and (12)(14) = 168, and both solutions are consecutive even integers, the conditions of the problem are met.

Therefore, the two integers are – 14 and – 12 or 12 and 14.

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Learner’s Material – Mathematics IX, First Edition pp. 27 - 34Holiday, Berchie. et. al. ALGEBRA 2. USA. The McGraw – Hill Companies, C2008. pp. 253 – 256 Wesner, et. al. ELEMENTARY ALGEBRA with APPLICATIONS. Bernard J. Klein Publishing, 2006 pp. 152 – 156

Reference Card

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Activity # 1 Quadratic or Not Quadratic? 1. NQE2. QE3. QE4. NQE5. QE

Answer Card

Activity # 2 Set Me to Your Standard1. - 2x2 + 3x – 7 = 0/2x2 – 3x + 7 = 02. - 2r2 – 6r + 5 = 0/2r2 + 6r – 5 = 03. 2x2 – 6x – 15 = 04. x2 + 7x + 12 = 05. x2 + 8x + 24 = 0

Activity # 3 What Made Me?

x2 + bx + c m + n m(n) (x + m) (x + n)

w2 – 8w + 12- 6 +

2 -6(2) (w – 6)(w + 2)

x2 + 5x – 248 + (-

3) 8(-3) (x + 8)(x – 3)

c2 + 6c + 5 5 + 1 5(1) (c + 5)(c + 1)

r2 + 5r – 147 + (-

2) 7(-2) (r + 7)(r – 2)

x2 + 9x + 20 5 + 4 5(4) (x + 5)(x + 4)

Assessment1. x2 + 8x + 16 = 0 (x + 4)(x + 4) = 0

x + 4 = 0x + 4 – 4 = 0 – 4 x = - 4

2. x2 – 5x – 14 = 0 (x – 7)(x – 2) = 0x – 7 = 0 x – 2 = 0x – 7 + 7 = 0 + 7 x – 2 + 2

= 0 + 2x = 7 x = 2

3. y2 + 9y + 20 = 0 (y + 5)(y + 4) = 0y + 5 = 0 y + 4 = 0 y + 5 –

5 = 0 – 5 y + 4 – 4 = 0 – 4 y = - 5 y = - 4

4. b2 – 10b + 21 = 0 (b – 7)(b – 3) = 0b – 7 = 0 b – 3 = 0b – 7 + 7 = 0 + 7 b – 3 + 3

= 0 + 3b = 7 b = 3

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