Pythagorean Quadratic Essay Example
Even though Ahmed’s half of the map doesn’t indicate which direction the 2x + 6 paces should go, we can assume that his and Vanessa’s paces should end up in the same place. I sketched this out on scratch paper I saw that it forms a right triangle with 2x + 6 being the length of the hypotenuse, and x and 2x + 4 being the legs of the triangle. Now I know how I can use the Pythagorean Theorem to solve for x.
The Pythagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse c, these lengths have the formula of a2 + b2 = c2.
Let a = x, and b = 2x + 4, so that c = 2x + 6.
Then, by putting these measurements into the Theorem equation we have x2 + (2x + 4)2 = (2x + 6)2 The binomials into the Phythagorean Therom x2 + 4x2 + 16x + 16 = 4x2 + 24x + 36 are the binomials squared. This a 4x2 on both sides of the equation which can be (-4x2 -4x2) subtracted out first leaving the equation to be x2 + 16x + 16 = 24x + 36. Next we should Subtract 16x from both sides of equation, which then leaves us with: x2 +16 = 8x + 36. The next step would then be to subtract 36 from both sides to get a result of. x2 -20= 8x. Finally we need to subtract 8x from both sides to get x2 – 8x – 20 =0. Now we have a quadratic equation to solve by factoring and using the zero factor.
(x – ) (x + ) = 0 Since the coefficient of x2 is 1 we can start with a pair of parenthesis with an x in each. Since the 20 is negative we know there will be one + and one – in the binomials. I noticed that in order for a number to multiply to -20 and add up to -8 the numbers would then have to be -10 and 2. So when I put that into the pair I ended up with (x – 10)(x + 2) = 0 Use the zero factor property to solve each binomial. Then I set each equation up to zero which made, x– 10 = 0 or x + 2 = 0 creating a compound equation.
When I did the math and solved for each problem, I got the answer of x = 10 or x = -2. These are the possible solutions to our equation. However, one
The Pythagorean Theorem states that in every right triangle with legs of length a and b and hypotenuse c, these lengths have the formula of a2 + b2 = c2.
Let a = x, and b = 2x + 4, so that c = 2x + 6.
Then, by putting these measurements into the Theorem equation we have x2 + (2x + 4)2 = (2x + 6)2 The binomials into the Phythagorean Therom x2 + 4x2 + 16x + 16 = 4x2 + 24x + 36 are the binomials squared. This a 4x2 on both sides of the equation which can be (-4x2 -4x2) subtracted out first leaving the equation to be x2 + 16x + 16 = 24x + 36. Next we should Subtract 16x from both sides of equation, which then leaves us with: x2 +16 = 8x + 36. The next step would then be to subtract 36 from both sides to get a result of. x2 -20= 8x. Finally we need to subtract 8x from both sides to get x2 – 8x – 20 =0. Now we have a quadratic equation to solve by factoring and using the zero factor.
(x – ) (x + ) = 0 Since the coefficient of x2 is 1 we can start with a pair of parenthesis with an x in each. Since the 20 is negative we know there will be one + and one – in the binomials. I noticed that in order for a number to multiply to -20 and add up to -8 the numbers would then have to be -10 and 2. So when I put that into the pair I ended up with (x – 10)(x + 2) = 0 Use the zero factor property to solve each binomial. Then I set each equation up to zero which made, x– 10 = 0 or x + 2 = 0 creating a compound equation.
When I did the math and solved for each problem, I got the answer of x = 10 or x = -2. These are the possible solutions to our equation. However, one