# Pythagorean Quadratic

**Topics:**Pythagorean theorem, Triangle, Right triangle

**Pages:**6 (726 words)

**Published:**September 22, 2014

Pythagorean Quadratic

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MAT 222 Introduction to Algebra

Instructor Yvette Gonzalez-Smith

August 04, 2013

Pythagorean Quadratic

The Pythagorean Theorem is an equation that allows a person to find the length of a side of a right triangle, as long as the length of the other two sides is known. The theorem basically relates the lengths of three sides of any right triangle. The theorem states that the square of the hypotenuse is the sum of the squares of the legs. It also can help a person to figure out whether or not the triangle is a right triangle or not, as long as the length of the other two sides are given (The Pythagorean Theorem, 1991-2012). This week’s assignment was found on page 371; it is problem number 98, called “buried treasure”. 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? Although we were not given a direction for Ahmed’s half of the map, we figure that his and Vanessa’s paces will end up in the same place. I have drawn a diagram on a piece of scratch paper, which includes the location of Castle Rock, and the dimensions of a triangle with the given number of paces. According to the Pythagorean Theorem, right triangles with legs the length of a and b, have a hypotenuse of c, are related to one another as I will let a = x, b = 2x+4, and c = 2x+6. Now, I will use the theorem in order to solve the equation. I have plugged the binomials into the equation for the theorem.

Using the FOIL method, I have performed the necessary multiplication; squaring a number is the same as multiplying it by itself. Next, I have

combined like terms. Next, I have subtracted the -4x squared from both sides. Next, I will subtract 24x from both sides. Next, I have subtracted 36 from both sides.

Here I am left with a quadratic equation. I will use factoring in order to solve, as well as the zero factor. 1 is the coefficient of ; therefore, I have started with a pair of parentheses with an x in each of them. I now need two factors -20 that will add to -8. -20, 1; -5, 4; 10, -2 20, -1; 5,-4; -10, 2 -10∙2 = -20; -10 + 2 = -8, so these numbers fit. Here we can use the zero factor property in order to solve each binomial; we have created a compound equation. When we add 20 to the first equation, or or subtract 2 from the second equation, here are the possible solutions to the equation. Since the 2 is negative, it becomes an extraneous solution. Extraneous solutions do not work simply because you cannot measure negative paces, or negative...

References: Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY:

McGraw-Hill Publishing

Polynomials. (2013). Retrieved from http://www.mathsisfun.com/algebra/polynomials.html

The Pythagorean Theorem. (1991-2012). Retrieved August 4, 2013, from the Pythagorean

Theorem website: http://www.purplemath.com/index.htm

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