191 CHAPTER 1 Test Prep Concepts Examples In a right triangle, the shorter leg is 7 in. less than the longer leg, and the hypotenuse is 2 in. greater than the longer leg. What are the lengths of the sides? Let x = the length of the longer leg. x – 7 x + 2 x 1x - 722 + x2 = 1x + 222 Substitute into the Pythagorean theorem. x2 - 14x + 49 + x2 = x2 + 4x + 4 Square the binomials. x2 - 18x + 45 = 0 Standard form 1x - 1521x - 32 = 0 Factor. x - 15 = 0 or x - 3 = 0 Zero-factor property x = 15 or x = 3 Solve each equation. The value 3 must be rejected because the height would be negative. The lengths of the sides are 15 in., 8 in., and 17 in. Check to see that the conditions of the problem are satisfied. The height of an object projected upward from ground level with an initial velocity of 64 ft per sec is given by s = -16t2 + 64t. Find the time(s) that the projectile will reach a height of 56 ft. 5 6 = -16t2 + 64t Let s = 56. 0 = -16t2 + 64t - 56 Subtract 56. 0 = 2t2 - 8t + 7 Divide by -8. t = -1-82 {21-822 - 4122172 2122 Quadratic formula with a = 2, b = -8, c = 7 t = 8{28 4 Simplify. t ≈1.29 or t ≈2.71 Use a calculator. The object reaches a height of 56 ft twice—once on its way up (after 1.29 sec) and once on its way down (after 2.71 sec). 1.5 Applications and Modeling with Quadratic Equations Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of legs a and b is equal to the square of the length of hypotenuse c. a2 +b2 =c2 Height of a Projected Object The height s (in feet) of an object projected directly upward from an initial height of s0 feet, with initial velocity v0 feet per second, is given by s =−16 t2 +v 0 t +s0, where t is the number of seconds after the object is projected.
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