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Question and Solution Both Here

In the solution in the above link, how did they assume x sq to be zero. If x sq is a real no then why couldn't they take is as some -ve number as it is less than 0?

Arthur
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Arpan Panda
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  • $\sqrt{x}$ is a increasing function which only takes real $x$. –  Jul 28 '17 at 10:35
  • @Fightclub1995. It seems that the question is about $x^2+8$ – Claude Leibovici Jul 28 '17 at 10:36
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    You're welcome to try finding real numnbers $x$ where $x^2$ is less than $0$. Try a negative number and see what happens. Trial and error is an important way to learn mathematics (and sadly underrated in our school system, where errors are often seen as fails, and students are afraid to attempt anything because making mistakes is somehow wrong). – Arthur Jul 28 '17 at 10:36

2 Answers2

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Lets say $x = -1,$

then $(-1)^2 + 8 \\=1+8 \\=9$

which is greater than if $x = 0$

this is because $(-1) \times (-1) = 1$

Evitan Retla
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By x sq you mean $x^2$ ?

If $x \in \mathbb R$, then $x^2 \ge 0$, hence $x^2+8 \ge 8$.

Conclusion: $8 \le x^2+8$ for all $x \in \mathbb R$.

For $x=0$ we have $8 =x^2+8$. Hence

$ \min\{x^2+8: x \in \mathbb R\}=8$

Fred
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