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Quadratic Equations and Prime Numbers Essay

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. (e) 2x + 2 = -7. 5 (f) 2x = 5. 52x = -9. 5 X = 2. 75x = -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 = 34x + 4 = 5. 3 16x ^2 – 16 x = 124x + 4 = 5. 3 4x + 4 = -5. 3 16x^2 – 16x + 16 = 12 + 164x = 1. 324x = -9. 32 X = . 33X = -2. 33 Equation (c) x^2 + 12x – 64 = 0 X^2 + 12x = 64 2x + 12 = 20 4x^2 + 48x + 144 = 2562x + 12 = 20 2x + 12 = -20 4x^2 + 48x + 144 = 256 + 1442x = 82x = -32 x^2 + 48 + 144 + 400x = 4 x= -16 Equation (d) 2x^2 – 3x – 5 = 0 2x^2 – 3x = 52. 8x + 3 = 5. 38 or 5. 4 (rounded) 8x^2 – 12x = 202. 8x + 3 = 5. 382. 8x + 3 = -5. 4 8x^2 – 12x + 9 = 20 + 92. 8 x = 1. 702. 8x = -8. 4 8x^2 -12x + 9 = 29x = . 85x = -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.

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I chose to use the numbers 0, 5,8,10, and 13, that is two even numbers and two odd numbers along with zero. Here are my equations and the results of searching to see if I can produce a prime number. X^2 – x + 41 = X^2 – x + 41 = 0^2 – 0 + 41 = 8^2 – 8 + 41 = 0 – 0 + 41 = 41 (prime number)64 -8 + 41 = 97 (prime number) X^2 – x + 41 =X^2 – x + 41 = 5^2 – 5 + 41 =10^2 – 10 + 41 = 25 – 5 + 41 = 61 (prime number)100 – 10 + 41 = 131 (prime number) X^2 – X + 41 = 13^2 – 13 + 41 = 169 – 13 + 41 = 197 (prime number) My equations gave me all prime numbers for my answers. A prime number is best described as numbers that have only two factors.

I even tried a few other numbers and again my answers were all prime numbers. I think that it is really interesting that all my answers are prime numbers. I believed that the reason I was getting all prime numbers was because I was adding a prime number in my equation. I then changed the number 41 to 40 and my answers were all composite numbers. Composite numbers are numbers that have three or more factors. Therefore I feel I was correct in saying if you add a composite number then your answer is a composite versus adding a prime number to get a prime number answer.

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Quadratic Equations and Prime Numbers Essay
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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 coefficie
2018-10-23 14:39:15
Quadratic Equations and Prime Numbers Essay
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