If x is a real number, what is the smallest possible value of x sq + 8?

Question

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?

Answer 1

Lets say $x = -1,$

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

which is greater than if $x = 0$

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

Answer 2

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$