Pythagorean Quadratic

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Pythagorean Quadratic
MAT221 Introduction to Algebra Pythagorean Quadratic
Week five of this class has been a complete challenge for me, from start to finish. Trying to master everything that we have been taught over the five weeks has truly been a test. I know there are benefits to knowing these principals, however, it stresses me to think about having to use it in real life circumstances.
This problem involves using the Pythagorean Theorem to find distance between several points in our textbook (Dugopolski, 2012). Ahmed has half of a treasure map,which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x?
We need to look at the equation so we can know how far Ahmed will have to walk, which is 2x+6 paces from Castle Rock. Even though Vanessa’s half of the map does not indicate in which direction the 2x + 4 paces should go, it can be assumed that her’s and Ahmed’s paces should end up in the same place. When sketched on scratch paper, a right triangle is formed with 2x + 6 being the length of the hypotenuse, and x and 2x + 4 being the legs of the triangle. When a right triangle is involved, the Pythagorean Theorem helps solve for x.

The Pyhagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse c, these length have the relationship of a2 + b2 =c2. Let a = x, and b = 2x + 6, so that c = 2x + 4 Then, putting these measurements into the Theorem, the equation becomes: x2 + (2x + 6)2 = (2x + 4)2 The binomials into the Pythagorean Theorem. x2 + 4x2 + 16x + 16 = 4x2 + 24x + 36 The binomials squared. Notice there…...

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