RUNNING HEAD: Pythagorean Quadratic 1

Pythagorean Theorem

MAT221

Instructor Gregory Dlabach

May 26, 2014

RUNNING HEAD: Pythagorean Quadratic 2

Pythagorean Theorem

In the final week’s assignment, a word exercise involves buried treasure in the textbook.

(Dugopolski, 2012). In this exercise Ahmed has a half of a treasure map that indicates there is a

buried treasure in the desert that is “2x+6” paces from Castle Rock. The exercise also states that

Vanessa has the other half of the map and it states that in order to find the buried treasure we

must go to “Castle Rock”, walk “x” paces north, and then walk “2x+4” paces to the east. The

assignment is to put together both Ahmed and Vanessa’s map and determine “x”. In order to

solve this exercise the “Pythagorean Theorem” is used. The “Pythagorean Theorem” in

algebra states that in a right triangle the square of “a2” plus the square of “b2” is equal to the

square of “c2”. This is further demonstrated in the diagram below:

a2 + b2 = c2

First in the equation it is noted that Ahmed will have to walk “2x+6” paces from Castle

Rock. If Ahmed drew a diagram and used a line to point to Castle Rock it would be labeled as

point “A” which is also “2x+6” paces. With this being the radius in order to find the treasure its

point is at “C” on the diagram. Vanessa will label her point on the diagram at point “B” which is

“2x+4” paces. The diagram example is below:

X

2x+4

2x+6

RUNNING HEAD: Pythagorean Quadratic 3

The “Pythagorean Theorem” theory states that the sum of the square of one leg and the

square of another leg would be equal to that of the square of the hypotenuse. With this

information an equation of “a2+b2=c2” can be written. In this equation “a” will equal to “x”, “b”

will equal to “2x+6”, and “c” equals “2x+4”. The detailed breakdown of solving this equation is

as follows:

x2+ (2x+6)2= (2x+4)2 The substitutions are added into the Pythagorean

Theorem

x2+ (2x+6) (2x+6) = (2x+4)2 The expression is multiplied by itself two times

x2+ (4x2+2x×6+6×2x+6×6) – (2x+4)2Using the FOIL method the equation is multiplied

and like terms are combined

x2+ (4x2+24x+36)-((2x+4(2x+4)) Each term is multiplied in the first group then in the

second group using the FOIL method.

x2+ (4x2+24x+36) - (4x2-16x-16) Each term is multiplied by “-1”

x2+4x2+24x+36-4x2-16x-16 Parentheses are removed and combine like terms

52+24x+36-4x2-16x-16 Add and subtract all like terms

X 2 + 8 x + 36-16, Subtract

x2-8x-20 Quadratic formula is used to solve by factoring and

using the zero factor; a=1, b=8, c=20

(x-)(x+) = 0 The coefficient of x2 is “1” therefore use

parentheses with “x”.

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(1x – 10)(1x+2)=0 The result of the quadratic formula which is the

original equation of x2-8x-20

(x-10) (x+2) Zero factor of x2-8x-20 – Therefore “x=10” and

“x=2”

x=10 and x=2 Compound equations

x=10 Used in beginning equation for Ahmed

2(10) +6 = 20+6 =26 26 paces

2(10) +4=20+4=24 24 paces

Combining our solution together, it states that Ahmed will have to go 26 paces from Castle Rock

in order to find the buried treasure. Vanessa will be 10 paces north and then have 24 paces east

to go.

RUNNING HEAD: Pythagorean Quadratic 5

References

Dugopolski, M. (2012). Linear Equations and Inequalities in One Variable. In Elementary and

intermediate algebra (pp. 145-151). New York, NY: McGraw-Hill.