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**Aim: How to solve verbal problems with quadratic equations?**

Do Now: The square of a number decreased by 4 times the number equals 21. Find the number.

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**Quadratic Verbal Problems**

The square of a number decreased by 4 times the number equals 21. Find the number. Let n equal the number The square of the number - n2 Four times a number - 4n n2 - 4n = 21 1. Put in standard form n2 - 4n - 21 = 0 2. Factor (n - 7)(n + 3) = 0 3. Set each factor equal to zero and solve. n + 3 = 0 n = -3 n - 7 = 0 n = 7

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**Quadratic Verbal Problems**

The product of two consecutive, positive, even integers is 80. Find the integers. 8 n - first of the consecutive, positive, even integers 10 n + 2 the next consecutive, positive, even integer n(n + 2) = 80 1. Put in standard form n2 + 2n = 80 n2 + 2n - 80 = 0 2. Factor (n + 10)(n - 8) = 0 3. Set each factor equal to zero and solve. n - 8 = 0 n = 8 n + 10 = 0 n = -10

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**Quadratic Verbal Problems**

The base of a parallelogram measures 7 centimeters more than its altitude. If the area of the parallelogram is 30 square centimeters, find the measure of its base and the measure of its altitude. Area of parallelogram A = bh x(x + 7) = 30 x x = altitude (height) x2 + 7x = 30 x + 7 x + 7 = base x2 + 7x - 30 = 0 Area = 30 (x + 10)(x - 3) = 0 x + 10 = 0 x = -10 x - 3 = 0 x = 3 ht. x + 7 = 10 base

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**Quadratic Verbal Problems**

The length of each of a pair of parallel sides of a square is increased by 2 meters, and the length of each of the other two sides of the square is decreased by 2 meters. The area of the rectangle formed is 32 square meters. Find the measure of one side of the original square. Area of square A = s2 s = side of original = 6 s + 2 = new length of first pair of sides s – 2 = new length of second pair of sides A of rectangle = (s + 2) (s – 2) = 32 s2 – 4 = 32 = 32 (6 + 2) (6 – 2) 8 · 4 = 32 s2 = 36 s = ±6

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Lesson 11.2 Area of Parallelograms and Triangles.

Lesson 11.2 Area of Parallelograms and Triangles.

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