Mat 221 Wk 5
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?
Given this scenario the Pythagorean Theorem would be the strategy we use to solve for x.
I started off with the Pythagorean Theorem.
I then plugged the binomials into the Pyth. Thrm.
Next I moved (2+6)^2 to le left of the equation by subtracting (2x+6)^2 from both sides.
I then squared the expression
Next I foiled the expression by multiplying and combining like terms.
I multiplied -1 by each term inside the parentheses and then removed the parentheses around the expression (4x^2 + 16x + 16)
Because x^2 an 4^2 are like terms I added 4x^2 to x^2 to get 5x^2
Since 5x^2 and -4x^2 are like terms I added -4x^2 to 5x^2 to get x^2
Again I added like terms then subtracted 36 from16 to get -20 giving us a quadratic equation to solve using the zero factor.
A compound equation.
Set each of the factors of the left hand side of the equation equal to 0.
Finally these are the possible solutions to the equation.
A^2 + b^2 = c^2 a = x b = 2x + 4 c = 2x + 6 x^2 + (2x + 4)^2 = (2x + 6)^2 x^2 + (2x + 4)^2 – (2x + 6)^2 = 0 x^2 + (2x + 4)(2 x+ 4) – (2x + 6)^2 = 0 x^2 + (4x^2 + 16x + 16) - (2x+6)^2 = 0 x^2+(4x^2 + 16x + 16) – (4x^2 +24x +36) =0 x^2 + 4x^2 + 16x + 16 - 4x^2 - 24x – 36 = 0
5x^2 + 16x +16 -4x^2 -24x -36 = 0
X^2 + 16x + 16 – 24x -36 = 0
X^2 – 8x +16 – 36 = 0
X^2 -8x – 20 = 0
(x+2)(x-10)=0
X+2 = 0 or x-10 = 0
X=-2, 10
However one of these solutions is irrelevant because it doesn’t work with this scenario at all. Since we cannot have negative paces or distance in a measured figure the solution -2 does not work. This leaves me with
Given this scenario the Pythagorean Theorem would be the strategy we use to solve for x.
I started off with the Pythagorean Theorem.
I then plugged the binomials into the Pyth. Thrm.
Next I moved (2+6)^2 to le left of the equation by subtracting (2x+6)^2 from both sides.
I then squared the expression
Next I foiled the expression by multiplying and combining like terms.
I multiplied -1 by each term inside the parentheses and then removed the parentheses around the expression (4x^2 + 16x + 16)
Because x^2 an 4^2 are like terms I added 4x^2 to x^2 to get 5x^2
Since 5x^2 and -4x^2 are like terms I added -4x^2 to 5x^2 to get x^2
Again I added like terms then subtracted 36 from16 to get -20 giving us a quadratic equation to solve using the zero factor.
A compound equation.
Set each of the factors of the left hand side of the equation equal to 0.
Finally these are the possible solutions to the equation.
A^2 + b^2 = c^2 a = x b = 2x + 4 c = 2x + 6 x^2 + (2x + 4)^2 = (2x + 6)^2 x^2 + (2x + 4)^2 – (2x + 6)^2 = 0 x^2 + (2x + 4)(2 x+ 4) – (2x + 6)^2 = 0 x^2 + (4x^2 + 16x + 16) - (2x+6)^2 = 0 x^2+(4x^2 + 16x + 16) – (4x^2 +24x +36) =0 x^2 + 4x^2 + 16x + 16 - 4x^2 - 24x – 36 = 0
5x^2 + 16x +16 -4x^2 -24x -36 = 0
X^2 + 16x + 16 – 24x -36 = 0
X^2 – 8x +16 – 36 = 0
X^2 -8x – 20 = 0
(x+2)(x-10)=0
X+2 = 0 or x-10 = 0
X=-2, 10
However one of these solutions is irrelevant because it doesn’t work with this scenario at all. Since we cannot have negative paces or distance in a measured figure the solution -2 does not work. This leaves me with