mat 117 week 7 DQ's 4 different explanation
mat 117 week 7 DQ's 4 different explanation
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Date Created: 11/16/15
Week 7 DQ 1 Due Day 2 Please post a 150 300 word response to the following discussion question by clicking on Reply. How do you know if a quadratic equation will have one, two, or no solutions? How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide your classmate’s with one or two solutions with which they must create a quadratic equation. Consider responding to your classmates by creating an equation from their solutions. Show that your equation should yield the appropriate number of solutions. If other equations exist with the same solution, provide the alternate equation and provide an explanation. You may want to view responses to the solutions you posted and guide your classmate’s if necessary. Explanation 1 How do you know if a quadratic equation will have one, two, or no solutions? The square root of a positive number is also some positive number. So in the numerator of the quadratic formula we will get two values: (b + the square root) and (b the square root). So when we get two solutions. B^24ac is zero. The square root of zero is zero. So in the numerator we get (b + 0) and (b 0). But both of these are equal to b.So when b^2 4ac we only get one solution. If the equation is negative then you will have no solution because you cannot square any real number and get a negative number. How do you find a quadratic equation if you are only given the solution? One way to find solutions from the equation is to factor it. For example, solving x^2 + 5x + 6 factor it: (x2)(x3) x2 = 0 or x3 =0 Solving these we get: x=2 or x=3 Is it possible to have different quadratic equations with the same solution? Yes it is possible to have to different equations with the same solution.For example Solution: x = 1 or x = 6 Equation: (x1)(x6) = 0 which gives x^2+7x+6 Explanation 2 A quadratic equations will have: one solution = the vortex, or like the very tip of the curve(where tangent = 0) intercepts with the x axis 2 solutions = where the formula intercepts the x axis at two positions 0 solution = where the curve doesn't touch the x axis An example to find the quadratic equation if only given one solution is x^2 + 6x + 9 = 0 (x+3)(x+3) =0 x = -3 Is it possible to have the same solution for two equations. example ax^2 + bx +c =y and ax^2 - bx -c = y I really am not comfortable with this question or making an equation to solve. So i will wait until i see some more examples before I try it. Explanation 3 To determine if a quadratic equation will have one, two, or no solutions, we must first find the discriminant. If b^24ac=0, then there is only one solution. If it is >0, then there are two. If it is <0, then there are no solutions. To find a quadratic equation when only the solution is given, you can recreate the equation as follows; y = k(xa)(xb). To recreate the equation, all one needs to do is replace the variables a and b with the solutions that are given. Yes, it is possible for two different quadratic equations to have the same solution because the variable k in y=k(xa)(xb) can have different values. Examples: x=4, x=3 x=5, x=7 Explanation 4 How do you know if a quadratic equation will have one, two, or no solutions? By looking at the discriminant D = b^2 - 4c: When D > 0, P(x) has two distinct real roots. When D = 0, P(x) has two coincident real roots. When D < 0, P(x) has no real roots. How do you find a quadratic equation if you are only given the solution? Let's say we are given the roots x = a and x = -b, then the quadratic equation is (x-a)(x+b) = 0. Is it possible to have different quadratic equations with the same solution? Yes, it is! Again, if we have the roots x = a and x = -b, then we can have C(x-a)(x+b) = 0, where C is a constant. So multiplying with a constant does not change the roots. Provide your classmate's with one or two solutions with which they must create a quadratic equation. Find a quadratic equation that has roots x = 2 and x = 5: (x-2)(x-5) = 0. Find a quadratic equation that has two roots at x = 1: (x-1)(x-1) = 0. Week 7 DQ 2 Due Day 4 Please post a 150 300 word response to the following discussion question by clicking on Reply. Quadratic equations may be solved by graphing, using the quadratic formula, completing the square, and factoring. What are the pros and cons of each of these methods? When might each method be most appropriate? Which method do you prefer? Explain why. Explanation 1 If graphing is used to solve a quadratic equation, you have to choose a number of values for x-coordinates to calculate the y-coordinates or vice versa. This process can take a lot of time and it is hard to determine what your solution is by looking at the graph if the solution is not a whole number. (Example x=1.2, x=-0.67) If the quadratic formula is used to solve a quadratic equation, this method can be used to solve any quadratic equation. It gives solution whether the solution has simple or complex numbers. Again process can be lengthy. If completing the square is used to solve a quadratic equation, this method can used for solving and quadratic but may sometimes be challenging if the quadratic does not have a perfect square binomial. (Example: x^2 + 14x = 4) Factoring can be used to solve quadratic only if the equation is factorable. (Example 6x^2 – 15x = 0) My first method for solving quadratic equations will be Factoring, only if the equation factors easily. My second method will be using the Quadratic formula if the equation cannot be factored. Explanation 2 Solving quadratic equations by graphing is probably the most lengthy and difficult process because you have to find the variables to graph. It does, however, give you a visual and if the line is straight then you know you did the problem correctly. Solving it by completing the square is also lengthy but efficient because you can check your work by plugging in the square roots you came up with. Factoring is the easiest option and most used because you complete the process by factoring the variables in. I prefer factoring over the rest of the options because I am not going to waste my time doing a graph if I don't have too (I also don't have the materials to do so lying around) and completing the square is almost backwards to me. None of these options are wrong to find the correct answer, it just depends on individual preference. Explanation 3 The single sure and easy way to solve them is using the quadratic formula. Just substitute, do a wee bit of arithmetic and you're done. It's main downside is that sometimes the answer can be seen a little more quickly with factoring. Factoring is second best. Some quadratics have factors that just jump out at you, but the best part about factoring is that when they do just jump out at you, most people kind of feel they got the answer for "free" (without having to work for it) and folks just love that feeling. For that reason it doesn't SEEM as hard as using the Quadric formula and doing the wee bit of arithmetic does. Folks like the feeling of easy. It's main downfall is that outside of tests and homework, the world is filled with Quadratic formulas that are HARD HARD HARD to factor. Completing the square is fairly pointless as it is harder than just using the Quadratic formula. It's error prone because you have to remember several steps and get them right. It's pretty pointless. And it's kind of just a specialized factoring approach. Graphing is a last resort. It is handy for when you have little to work with (perhaps the information is buried in a word problem) or just need an approximate answer. The Quadratic formula is absolutely the best, except for the simplest and most memorable of equations. I mean, if I have to factor x^2 1, I'm doing it with factoring, but only because I've known the answer for decades. Anything that doesn't just jump out, I'm using the Quadratic formula. Explanation 4 A pro concerning graphing is that you can quickly input numbers into a graphing calculator to get an answer. However, the con of this is that the answer you get is a roundabout answer. A pro concerning the quadratic formula is that the process is very exact. The con of this would be that the process, although exact, may take a while to arrive at your answer. A pro of completing the square is that it allows you to derive precise answers. A con of this can be the amount of time it takes to solve using this method. A pro of factoring is that it is the simplest method in my opinion. A con of factoring would be how it may become confusing over time if you are needing to find answers for many different variables. I would have to say my favorite method is factoring. Factoring seems to fit my style of completing any algebra problem. Plus, it makes you use your brain in a different way in my opinion.
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