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Comparing numbers....Grade level 10

Compare the numbers:
\(\sqrt{x^2+1}\) and \(x\)

Note by Oussama Jaber
3 years, 8 months ago

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I am assuming that the principle square root is taken. Note that \(x=\sqrt{x^2}\). Obviously \(\sqrt{x^2+1}>\sqrt{x^2}\); therefore \(\sqrt{x^2+1}>x\).

CAUTION: observe that \(\sqrt{x^2}=|x|\ne x\); however, even if \(x\) is negative it does not invalidate the inequality. Daniel Liu · 3 years, 8 months ago

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