## College Algebra (10th Edition)

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 $x = 13$: $$Hypotenuse = 2(13) + 3 = 29$$$$Leg1 = 2(13)-5 = 21$$$$Leg2 = (13)+7 = 20$$