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Solve Application problem using quadratic equation?

Solve: A landscaper is designing a flower garden in the shape of a right triangle. She wants 10ft of a perennial border to form the hypotenuse of the triangle, and one leg is to be 2ft longer than than the other. Find the width of the shortest leg. A. 13ft B. 6ft. C. 5.742ft D. 8ft. E. 1.742ft

Public Comments

1. Let "w" be the width of the shortest leg. Then the other leg has to be w+2. The hypotenuse is 10, so the Pythagorean Theorem says: x^2 + (x+2)^2 = 10^2 Expand this and move everything over to the left. This will give you an equation in the form of ax^2 + bx + c = 0, which you can then solve. You may end up with two solutions, one of which is negative. Obviously the negative won't apply, since you can't have a border in "negative feet".
2. let x = shorter leg, then longer leg is x+2 by Pythagorean theorem: x^2 + (x+2)^2 = 10^2 expand the binomial: x^2 + x^2 + 4x + 4 = 100 write in standard form: 2x^2 + 4x - 96 = 0 (I like to divide out common 'constant' factors) x^2 + 2x - 48 = 0 use quadratic formula to get x = 6 or x = -8 toss -8 as lengths are measured in positives so answer B: 6 ft
3. If a is the shorter leg and b is the longer leg... a²+b²=10² Since b=a+2... a²+(a+2)² = 100 a²+a²+4a+4 = 100 2a²+4a+4=100 2a²+4a-96=0 2(a²+2a-48) = 0 2(a-6)(a+8) = 0 a=6, a=-8 Since we're trying to find the leg of a triangle, a=-8 doesn't make sense. So a=6. The answer is B.