Quadratic Equations and Prime Numbers Essay Example
Solving the quadratic equations using the FOIL method makes the equations easier for me to understand. The Foil method, multiplying the First, Outer, Inner and Last numbers, breaks down the equation a little further so you understand where some of your numbers are coming from, plus it helps me to check my work.
Equation (a.) x^2 – 2x – 13 = 0 X^2 – 2x = 13 (step a) 4x^2 – 8x = 52 (step b, multiply by 4) 4x^2 – 8x + 4 = 52 + 4 (step c, add to both sides the square of original coefficient) 4x^2 – 8x + 4 = 56 2x + 2 = 7.5 (d, square root of both sides) 2x + 2 = 7.5 (e) 2x + 2 = -7.5 (f) 2x = 5.5 2x = -9.5 X = 2.75 x = -4.75 (2x +2) (2x +2) 2x X 2x = 4x^2 (foil method) 2x X 2 = 4x 2 x 2x = 4x 2 x 2 = 4 Simplify it 4x^2 – 8x + 4
Equation (b) 4x^2 – 4x + 3 = 0 16x^2 -16x + 16 = 28 4x^2 – 4x = 3 4x + 4 = 5.3 16x ^2 – 16 x = 12 4x + 4 = 5.3 4x + 4 = -5.3 16x^2 – 16x + 16 = 12 + 16 4x = 1.32 4x = -9. 32 X = .33 X = -2.33
Equation (c) x^2 + 12x – 64 = 0 X^2 + 12x = 64 2x + 12 = 20 4x^2 + 48x + 144 = 256 2x + 12 = 20 2x + 12 = -20 4x^2 + 48x + 144 = 256 + 144 2x = 8 2x = -32 4x^2 + 48 + 144 + 400 x = 4 x= -16
Equation (d) 2x^2 – 3x – 5 = 0 2x^2 – 3x = 5 2.8x + 3 = 5.38 or 5.4 (rounded) 8x^2 – 12x = 20 2.8x + 3 = 5.38 2.8x + 3 = -5.4 8x^2 – 12x + 9 = 20 + 9 2.8 x = 1.70 2.8x = -8.4 8x^2 -12x + 9 = 29 x = .85 x = -3
I really got the hang of these equations by equation d and started enjoying figuring them out. I think the India method is an interesting method to solve equations and for me, I could understand it easier then some of the other methods we have been using.
Using the formula X^2 –X + 41 to try if we can get a prime number was fun and interesting. I chose to use the numbers 0, 5,8,10, and 13, that is two even numbers and two
Equation (a.) x^2 – 2x – 13 = 0 X^2 – 2x = 13 (step a) 4x^2 – 8x = 52 (step b, multiply by 4) 4x^2 – 8x + 4 = 52 + 4 (step c, add to both sides the square of original coefficient) 4x^2 – 8x + 4 = 56 2x + 2 = 7.5 (d, square root of both sides) 2x + 2 = 7.5 (e) 2x + 2 = -7.5 (f) 2x = 5.5 2x = -9.5 X = 2.75 x = -4.75 (2x +2) (2x +2) 2x X 2x = 4x^2 (foil method) 2x X 2 = 4x 2 x 2x = 4x 2 x 2 = 4 Simplify it 4x^2 – 8x + 4
Equation (b) 4x^2 – 4x + 3 = 0 16x^2 -16x + 16 = 28 4x^2 – 4x = 3 4x + 4 = 5.3 16x ^2 – 16 x = 12 4x + 4 = 5.3 4x + 4 = -5.3 16x^2 – 16x + 16 = 12 + 16 4x = 1.32 4x = -9. 32 X = .33 X = -2.33
Equation (c) x^2 + 12x – 64 = 0 X^2 + 12x = 64 2x + 12 = 20 4x^2 + 48x + 144 = 256 2x + 12 = 20 2x + 12 = -20 4x^2 + 48x + 144 = 256 + 144 2x = 8 2x = -32 4x^2 + 48 + 144 + 400 x = 4 x= -16
Equation (d) 2x^2 – 3x – 5 = 0 2x^2 – 3x = 5 2.8x + 3 = 5.38 or 5.4 (rounded) 8x^2 – 12x = 20 2.8x + 3 = 5.38 2.8x + 3 = -5.4 8x^2 – 12x + 9 = 20 + 9 2.8 x = 1.70 2.8x = -8.4 8x^2 -12x + 9 = 29 x = .85 x = -3
I really got the hang of these equations by equation d and started enjoying figuring them out. I think the India method is an interesting method to solve equations and for me, I could understand it easier then some of the other methods we have been using.
Using the formula X^2 –X + 41 to try if we can get a prime number was fun and interesting. I chose to use the numbers 0, 5,8,10, and 13, that is two even numbers and two