College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 93

Answer

Two right triangles exist that satisfy the conditions established: One with dimensions (13, 12, 5) and another one with dimensions (29, 21, 20).

Work Step by Step

To analyze the problem, it helps to draw the triangle as described in the exercise (refer to the figure provided). We know that all sides of a right triangle are related through the Pythagorean Theorem: $a^{2} + b^{2} = c^{2}$ where $c$ represents the hypotenuse and $a,b$ represent the legs. If we were to substitute the values provided in the exercise into the Pythagorean Theorem, we would obtain the following equation: $$(2x-5)^{2} + (x+7)^{2} = (2x + 3)^{2}$$ which we can expand into: $$(4x^{2} - 20x + 25) + (x^{2} + 14x + 49) = (4x^{2} + 12x + 9)$$ Re-writing the equation so that it equals to zero results in the following: $$4x^{2} - 20x + 25 + x^{2} + 14x + 49 - 4x^{2} - 12x - 9 = 0 $$$$x^{2} - 18x +65 = 0$$This equation can be solved by multiple methods. Factorizing is one of them, where we would be looking for two factors of 65 that, when added, give -18. These factors are -5 and -3. Therefore, we can factorize, and solve, the equation: $$x^{2} - 18x +65 = 0$$$$(x - 13)(x-5) = 0$$$$(x - 13) = 0$$$$x = 13$$$$(x - 5) = 0$$$$x = 5$$ We now have two possible values for $x$ which we can apply to the given dimension expressions. For $x = 5$: $$Hypotenuse = 2(5) + 3 = 13$$$$Leg1 = 2(5)-5 = 5$$$$Leg2 = (5) + 7 = 12$$ For $x = 13$: $$Hypotenuse = 2(13) + 3 = 29$$$$Leg1 = 2(13)-5 = 21$$$$Leg2 = (13)+7 = 20$$
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