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Bell Work: Find the values of all the unknowns: R T = R T T + T = 60 R = 3 R = 6 1 1 2 2 1 2 1 2

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Answer: T = 20 T = 40 2121

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Lesson 88: Quadratic Equations, Solution of Quadratic Equations by Factoring

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Quadratic equations are second degree polynomial equations. Second degree in x means that the greatest exponent of x in any term is 2. both of these equations are quadratic equations in x because the greatest exponent of x is 2. 4 – 3x = 2x3x – 2x + 4 = 0 2

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The first equation was in standard form because all nonzero terms are on the left of the equals sign and the terms are written in descending order of the variable. The coefficient of x cannot be zero, but either of the other two numbers can be zero. Thus, each of the following equations is also a quadratic equation in x: 4x = 0 4x + 2x = 0 4x – 3 = 0 2 2 2 2

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To designate a general quadratic equation, we use the letter a to represent the coefficient of x, the letter b to represent the coefficient of x, and the letter c to represent the constant term. Using these letters to represent the constants in the equation, we can write a general quadratic equation in standard form as ax + bx + c = 0 2 2

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If we let a = 1, b = -3, and c = -10, we have the equation x – 3x – 10 = 0 If we substitute either 5 or -2 for the variable x in the quadratic equation, the equation will be transformed into a true equation. If x = 5If x = -2 (5) – 3(5) – 10 = 0 (-2) – 2(-2) – 10 = 0 0 = 0 2 2

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The numbers 5 and -2 are the only numbers that will satisfy the equation before. Every quadratic equation has at most two distinct numbers that will make the equation a true statement.

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Some quadratic equations can be solved by using the zero factor theorem. Zero factor theorem: if p and q are any real numbers and if p x q = 0, then either p = 0 or q = 0, or both p and q equal 0.

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For example, (x – 3)(x + 5) = 0 Here we have two quantities multiplies and the product is equal to zero. From the zero factor theorem, we know that at least one of the quantities must equal zero if the product is to equal zero.

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So either x – 3 = 0x + 5 = 0 x = 3x = -5 Thus the two values of x that satisfy the condition stated are 3 and -5.

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We can use the zero factor theorem to help us solve quadratic equations that can be factored. We do this by first writing the equation in standard form and factoring the polynomial; then we set each of the factors equal to zero and solve for the values of the variable.

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Example: Use the factor method to find the roots of x – 18 = 3x. 2

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Answer: x = -3, 6

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Example: Find the roots of -25 = -4x 2

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Answer: x = 5/2, -5/2

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Example: Find the values of x that satisfy x – 56 = -x. 2

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Answer: x = -8, 7

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Example: Solve 3x – 6x = 9 2

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Answer: x = 3, -1

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HW: Lesson 88 #1-30

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