A quadratic equation is shown below: 9x2 - 36x + 36 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (3 points) Part B: Solve 2x2 - 9x + 7 = 0 using an appropriate method. Show the steps of your work and explain why you chose the method used.(4 points) Part C: Solve 3x2 - 12x + 2 = 0 by using a method different from the one you used in Part B.

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A quadratic equation is shown below: 9x2 - 36x + 36 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (3 points) Part B: Solve 2x2 - 9x + 7 = 0 using an appropriate method. Show the steps of your work and explain why you chose the method used.(4 points) Part C: Solve 3x2 - 12x + 2 = 0 by using a method different from the one you used in Part B.

Algebra
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You have to use the quadratic equation for Part A (since you have to find the radicand for one of the solutions) Part A : Quadratic Equation \[x=-b \pm \sqrt{b ^{2}-4ac}\div2a\] you get b, a, and c through this format: (a)x^2 - (b)x + (c) = 0 So look at 9x2 - 36x + 36 = 0, and put them in the quadratic equation without the variable x.
For Part B and C, you can either use the quadratic equation again to find the solutions or the slip and slide method, since you have to use two different methods for each. The slip and slide is used when you have a coefficient on your first term. Let's use Part B for the slip and slide method: \[2x^2-9x+7=0\] 1) take the 2 from the 2x^2 and multiply it to the 7 leaving you with: \[x^2-9x+14\] 2) Begin to start finding the solutions like you would when your first term doesn't have a coefficient: \[(x-2)(x-7)\] 3) Solve for the two x's on the parentheses to find your solution: \[x-2+2=0+2\] \[x=2\] \[x-7+7=0+7\] \[x=7\] Your two solutions are x=7 and x=2

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