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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If I put 0 at the original equation, I have it correct, why 0 is not a root?
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blonpina
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If I put 0 at the original equation, I have it correct, why 0 is not a root?
The explanation is clear and elaborate enough.

If x = 0, then ­\(2-x^2 = (x-2)^2\) becomes:

­\(2-0^2 = (0-2)^2\)

­\(2 = (-2)^2\)

­\(2 = 4\)

The above is not correct. Hence, 0 is not a solution. 
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand understand how 2-x^2= x^2-4x+4 equals to x^2-2x+1.
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S1256
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand understand how 2-x^2= x^2-4x+4 equals to x^2-2x+1.

This is very basic algebraic manipulation.

\(2-x^2=x^2-4x+4\);

Subtract 2 - x^2 from both sides

\(0=x^2-4x+4 - (2-x^2) \);

\(0=2x^2-4x+2 \);

Reduce by 2:

\(0=x^2-2x+1 \).
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