Answer:
Step-by-step explanation:
Was going to get y’all there in
So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
![x-2=\sqrt{x-2}](https://tex.z-dn.net/?f=x-2%3D%5Csqrt%7Bx-2%7D)
Next, square both sides:
![(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2](https://tex.z-dn.net/?f=%28x-2%29%5E2%3Dx-2%5C%5C%28x-2%29%28x-2%29%3Dx-2%5C%5Cx%5E2-4x%2B4%3Dx-2)
Next, subtract x and add 2 to both sides of the equation:
![x^2-5x+6=0](https://tex.z-dn.net/?f=x%5E2-5x%2B6%3D0)
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
![x^2-2x-3x+6=0](https://tex.z-dn.net/?f=x%5E2-2x-3x%2B6%3D0)
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
![x(x-2)-3(x-2)=0](https://tex.z-dn.net/?f=x%28x-2%29-3%28x-2%29%3D0)
Now you can rewrite the equation as ![(x-3)(x-2)=0](https://tex.z-dn.net/?f=%28x-3%29%28x-2%29%3D0)
Now, apply the Zero Product Property and solve for x as such:
![x-3=0\\x=3\\\\x-2=0\\x=2](https://tex.z-dn.net/?f=x-3%3D0%5C%5Cx%3D3%5C%5C%5C%5Cx-2%3D0%5C%5Cx%3D2)
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
![2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}](https://tex.z-dn.net/?f=2%3D2%2B%5Csqrt%7B2-2%7D%5C%5C2%3D2%2B%5Csqrt%7B0%7D%5C%5C2%3D2%2B0%5C%5C2%3D2%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C3%3D2%2B%5Csqrt%7B3-2%7D%5C%5C3%3D2%2B%5Csqrt%7B1%7D%5C%5C3%3D2%2B1%5C%5C3%3D3%5C%20%5Ctextsf%7Btrue%7D)
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
Answer:
C
Step-by-step explanation:
C because x repeats
15
five three times is equal to fifteen
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