Solving Quadratic Equations Discussion - 250 words needed

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NAT164

Mathematics

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In this discussion, you will solve quadratic equations by two main methods: factoring and using the Quadratic Formula. Read the following instructions in order and view the example to complete this discussion:

Complete problem attached   Week 4 discussion instructions.pdf :   

A or L4648 on p. 646
  • For the factoring problem, be sure you show all steps to the factoring and solving. Show a check of your solutions back into the original equation.
  • For the Quadratic Formula problem, be sure that you use readable notation while you are working the computational steps. Refer to Inserting Math Symbols for guidance with formatting.
  • Present your final solutions as decimal approximations carried out to the third decimal place. Due to the nature of these solutions, no check is required.
  • Incorporate the following four math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):
    • Quadratic formula
    • Factoring
    • Completing the square
    • Discriminant
Your initial post should be at least 250 words in length.

Example attached MAT222.W4.DiscussionExample.pdf 

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INSTRUCTOR GUIDANCE EXAMPLE: Week Four Discussion Solving Quadratic Equations #79 pg 637: Solve by Factoring There is another way this problem could be solved (by completing the square which has already been done to it) but our instructions say to solve by factoring. This will require us to multiply out the left side and then subtract 9/4 from both sides to leave the right side zero. (p + ½ )2 = 9/4 First we need to expand the left side by FOIL. 2 p + p + ¼ = 9/4 Subtract 9/4 from both sides. p2 + p – 2 = 0 Since ¼ – 9/4 = -8/4 = -2 we now are free of the fractions. (p + 2)(p – 1) = 0 Left side is factored. p + 2 = 0 or p – 1 = 0 Using the Zero Factor Property. p = -2 or p = 1 Our solutions. {-2, 1} Solution set presented. Check: (p + ½ )2 = 9/4 (-2 + ½ )2 = 9/4 (-3/2)2 = 9/4 9/4 = 9/4 #87 pg 637 -x2 + x + 6 = 0 -1(x2 – x – 6) = 0 x2 – x – 6 = 0 (x – 3)(x + 2)= 0 x – 3 = 0 or x + 2 = 0 x = 3 or x = -2 {-2, 3} Check: -x2 + x + 6 = 0 -(-2)2 + (-2) + 6 = 0 -4 – 2 + 6 = 0 -6 + 6 = 0 0=0 (p + ½ )2 = 9/4 (1 + ½)2 = 9/4 (3/2)2 = 9/4 9/4 = 9/4 Factor -1 out of all terms first. Divide both sides by -1 Ready for factoring. Left side is factored. Using the Zero Factor Property. Our solutions. Solution set presented. -x2 + x + 6 = 0 -(3)3 + 3 + 6 = 0 -9 + 3 + 6 = 0 -6 + 6 = 0 0=0 #47 pg 646: Solve using Quadratic Formula 3y2 + 2y – 4 = 0 a = 3, b = 2, c = -4 Discriminant is b2 – 4ac which is 22 – 4(3)(-4) = 52 so we have two real solutions. y = -(2) √[22 – 4(3)(-4)] All values put into the formula in parenthesis. 2(3) y = -2 √[4 + 48] Simplification begins. 6 y = -2 √[52] Need to simplify the radical next: 52 = 4 13 6 y = -2 2√[13] 6 y = -1 √[13] 3 {.869, -1.535} Both terms in the top and 6 have a factor of 2 which can be canceled out. This is our solution set in radical form. Our solution set presented as decimal approximations. Using the Quadratic Formula will work on all types of quadratic equations, but factoring is quicker and easier if it is a possible choice for the equation.
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Anonymous
Excellent resource! Really helped me get the gist of things.

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