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Aim: Solving Verbal Problems using Quadratic Equations Course: Adv. Alg. & Trig 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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Aim: Solving Verbal Problems using Quadratic Equations Course: Adv. Alg. & Trig 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 - n 2 Four times a number - 4n n 2 - 4n = Put in standard form n 2 - 4n - 21 = 0 2. Factor (n - 7)(n + 3) = 0 3. Set each factor equal to zero and solve. n + 3 = 0n = -3 n - 7 = 0n = 7

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Aim: Solving Verbal Problems using Quadratic Equations Course: Adv. Alg. & Trig Quadratic Verbal Problems The product of two consecutive, positive, even integers is 80. Find the integers. n - first of the consecutive, positive, even integers n + 2 the next consecutive, positive, even integer n(n + 2) 1. Put in standard form n 2 + 2n = 80 n 2 + 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 = = 80

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Aim: Solving Verbal Problems using Quadratic Equations Course: Adv. Alg. & Trig 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. x x = altitude (height) x + 7 x + 7 = base Area of parallelogram A = bh x(x + 7) = 30 x 2 + 7x = 30 x 2 + 7x - 30 = 0 x + 10 = 0 x = -10 x - 3 = 0 x = 3 x + 7 = 10 ht. base Area = 30 (x + 10)(x - 3) = 0

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Aim: Solving Verbal Problems using Quadratic Equations Course: Adv. Alg. & Trig 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 = s 2 A of rectangle == 32 s = side of original s + 2 = new length of first pair of sides s – 2 = new length of second pair of sides (s + 2)(s – 2) s 2 – 4 = 32 s 2 = 36 s = ±6 = 6 = 32 (6 + 2)(6 – 2) 8 · 4 = 32

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