Presentation on theme: "Aim: How to solve verbal problems with quadratic equations?"— Presentation transcript:
1 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.
2 Quadratic Verbal Problems The square of a number decreased by 4 times the number equals 21. Find the number.Let n equal the numberThe square of the number - n2Four times a number - 4nn2 - 4n = 211. Put in standard formn2 - 4n - 21 = 02. Factor(n - 7)(n + 3) = 03. Set each factor equal to zero and solve.n + 3 = 0n = -3n - 7 = 0n = 7
3 Quadratic Verbal Problems The product of two consecutive, positive, even integers is 80. Find the integers.8n - first of the consecutive, positive, even integers10n + 2 the next consecutive, positive, even integern(n + 2)= 801. Put in standard formn2 + 2n = 80n2 + 2n - 80 = 02. Factor(n + 10)(n - 8) = 03. Set each factor equal to zero and solve.n - 8 = 0n = 8n + 10 = 0n = -10
4 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 = bhx(x + 7) = 30xx = altitude (height)x2 + 7x = 30x + 7x + 7 = basex2 + 7x - 30 = 0Area = 30(x + 10)(x - 3) = 0x + 10 = 0x = -10x - 3 = 0x = 3ht.x + 7 = 10base
5 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 = s2s = side of original= 6s + 2 = new length of first pair of sidess – 2 = new length of second pair of sidesA of rectangle =(s + 2)(s – 2)= 32s2 – 4 = 32= 32(6 + 2)(6 – 2)8 · 4 = 32s2 = 36s = ±6