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# M18-35

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Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626020 [11]
Given Kudos: 81940
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626020 [4]
Given Kudos: 81940
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626020 [0]
Given Kudos: 81940
Intern
Joined: 05 Feb 2019
Posts: 2
Own Kudos [?]: 0 [0]
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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?
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626020 [1]
Given Kudos: 81940
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blonpina wrote:
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.