Week 3 Written Assignment (Mat126)
Week Three Written Assignment mustangsallygt08 MAT126
February 18, 2013
Week Three Written Assignment The steps that are being followed to solve quadratic equations that came from India, and the steps are: (a) Move the constant tern to the right side of the equation. (b) Multiply each term in the equation by four times the coefficient of the x2 term. (c) Square the coefficient of the original x term and add it to both sides of the equation. (d) Take the square root of both sides. (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. (f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
Let’s solve the equation (a): x2 –2x – 13 = 0 x2 – 2x = 13
4x2 – 8x = 52
4x2 – 8x – 13 = 52 – 13
4x2 – 8x – 13 = 39
2x – 13 = 39
2x – 13 = 39 2x – 13 = -39
2x = 26 2x – 13 = -26 x=13 x = -13
Now that the steps were used and were easy to follow and understand let’s apply them to equation (c): x2 + 12x – 64 = 0 x2 + 12x = 64
4x2 + 48x = 248
4x2 + 48x – 64 = 248 – 64
4x2 + 48x – 64 = 184
3x – 2 = 23 3x-2=23 3x – 2 = -23
3x = 21 3x = -21 x = 7 x = -7
This can be handy when you need to solve a quadratic equation in real life, and I could use it at work to find out the lowest price that certain steaks or roast can be sold at while the company is still making a nice profit to insure a raise in my
February 18, 2013
Week Three Written Assignment The steps that are being followed to solve quadratic equations that came from India, and the steps are: (a) Move the constant tern to the right side of the equation. (b) Multiply each term in the equation by four times the coefficient of the x2 term. (c) Square the coefficient of the original x term and add it to both sides of the equation. (d) Take the square root of both sides. (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. (f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.
Let’s solve the equation (a): x2 –2x – 13 = 0 x2 – 2x = 13
4x2 – 8x = 52
4x2 – 8x – 13 = 52 – 13
4x2 – 8x – 13 = 39
2x – 13 = 39
2x – 13 = 39 2x – 13 = -39
2x = 26 2x – 13 = -26 x=13 x = -13
Now that the steps were used and were easy to follow and understand let’s apply them to equation (c): x2 + 12x – 64 = 0 x2 + 12x = 64
4x2 + 48x = 248
4x2 + 48x – 64 = 248 – 64
4x2 + 48x – 64 = 184
3x – 2 = 23 3x-2=23 3x – 2 = -23
3x = 21 3x = -21 x = 7 x = -7
This can be handy when you need to solve a quadratic equation in real life, and I could use it at work to find out the lowest price that certain steaks or roast can be sold at while the company is still making a nice profit to insure a raise in my