PP. 388-393 SOLVING EQUATIONS 4.8(EX. 1-3,7,8) BY USING ... session 10 4.8(ex. 1-3,7,8) pp. 388-393...
Transcript of PP. 388-393 SOLVING EQUATIONS 4.8(EX. 1-3,7,8) BY USING ... session 10 4.8(ex. 1-3,7,8) pp. 388-393...
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SESSION 104.8(EX. 1-3,7,8)PP. 388-393
SOLVING EQUATIONS BY USING THE ZERO PRODUCT RULE
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Quadratic equation
▪Definition: Quadratic equation in one variable is an equation that can be written in the form: ax² + b x +c = 0 a≠0
▪& a , b , c are real numbers
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Write into the standard form
▪ Check if these equations are quadratic:
▪ 1) -4 x² +4x=1▪ 2) x(x-2)=3▪ 3)(x-4)(x+4)=0
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Using the product rule to solve the equation
The zero product rule:If a product a •b =0 then a=0 or b= 0So if let say: (x-4)(x+4)=0 then x-4=0 or x+4=0 x=4 or x=-4
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Ex.1. 2x²-5x = 12▪First simplify:2x²-5x -12=0▪AC product 2∙(-12)=-24 ▪Sum; -5 (what is the strategy to find the numbers).
▪Numbers are: -8, 5
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2x²-5x = 12
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2x²-5x -12=0▪ 2x²-8x + 3x-12=0▪ (2x²-8x) + (3x-12)=0▪ 2x (x-4) + 3(x-4)=0▪ (X-4)(2x+3)=0 (0 pr rule)▪ X-4=0 or 2x+3=0▪ X=4 x=- 3/2 (-1.5)
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Ex.2. 6x²+8x=0
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Ex.2. 6x²+8x=0
▪2x(3x+4)=0▪X=0 or 3x+4=0▪ 3x=-4▪ x=-4/3▪Check the answer.
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Ex.3. 9x(4x+2)-10x=8x+25
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Ex.3. 9x(4x+2)-10x=8x+25
▪36x²+18x-10x=8x+25▪36x²+8x= 8x+25▪36x²+8x- 8x-25=0▪36x²-25=0▪(6x)²-5²=0▪(6x-5)(6x+5)=0▪X=5/6 or x=-5/6
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Ex.7.The product of two consecutive odd integers is 35. Find integers.
▪First odd integer x▪Second odd integer will be x+2
▪Product: x(x+2)=35▪Solve it for x. & find the consecutive numbers.
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Application using the quadratic equation
▪ The length of a basketball court is 6ft less than 2 times the width. If the total area is 4700 ft ², find the dimensions of the court.
▪ w
L
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Application using the quadratic equation
▪ The length of a basketball court is 6ft less than 2 times the width. If the total area is 4700 ft ², find the dimensions of the court.
▪ w
2w-6A=L•W 4700= (w-6)•w Solve it for w
that find the L w=50 (w=-47)
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ax²+bx+c=0 where ax²+bx+c is a perfect square
▪4x²+8x+4= 0
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4x²+8x+4
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4x²+8x+4=0Factor first Use the formula first
▪ Method 1:▪ 4(x²+2x+1)=0▪ 4(x+1)²=0▪ X+1=0▪ X=-1
▪ 4x²+8x+4=0▪ (2x+2) ²=0▪ 2x+2=0▪ 2(x+1)=0▪ x+1=0▪ X=-1
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Solve equations by using the zero product rule▪x²-2x -24=0
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x²-2x -24=0
▪x²-2x -24=0▪(x-6)(x+4)=0▪x-6=0 or x+4=0
▪So: x=6 or x=-4
• Using AC rule to factor
• Pr. -24, sum -2• Numbers: -6, 4• Use zero product rule
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9x²-12x=0
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9x²-12x=0▪ 9x²-12x=0▪ 3x ( 3x-4)= 0▪ 3x=0 or 3x-4=0▪ X=0 or x=4/3▪ Check:▪ 9•0²-12•0=0▪ 9•(4/3)²-12•(4/3)=0▪ 0=0
▪ List the strategies to factor:
▪ Look for common factor
▪ Look for the number of terms (2 terms, 3 terms etc
▪ Look for formulas
▪ If a trinomial and the formulas don’t work use AC method.
▪ Factor completely
▪ Check by multiplying
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3x(2x-1)-x=2x(x-2)+25
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Simplify First▪ 3x(2x-1)-x=2x(x-2)+25▪ 6x²-3x-x = 2x²-4x +25▪ 6x²-4x = 2x²-4x +25▪ 6x²-2x²-4x+4x -25=0▪ 4x²-25=0▪ (2x-5)(2x+5)=0▪ X=5/2 or x=-5/2 (2.5)
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Solving higher-degree polynomial equation▪z³+3z²-4z-12=0
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Solving higher-degree polynomial equationz³+3z²-4z-12=0(z³+3z²)-(4z+12)=0 (group) z²(z+3)- 4(z+3)=0 (z+3)(z²-4)=0 (z+3)(z-2)(z+2) =0 Z=-3; z=2; z=-2
HW #10
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Summarizing of the session
Solving equations by zero product rule▪ * Simplify the equation in the standard
form ▪ ax²+bx +c=0▪ Factor completelyApply the product rule by equaling to zero
each factor and solving the equations that you get.
▪ For the higher degree equations the rules work almost the same.