Use examples to explain why $\sqrt x \cdot \sqrt x=-x$ for all negative values of $x$

Question

I am helping my daughter with this problem in Saxon Algebra $1$ (Lesson $69$) and came across this question.

Use examples to explain why $\sqrt x \cdot \sqrt x=-x$ for all negative values of $x$.

I think when $x$ is negative, $\sqrt x$ is not defined and the question does not make sense. It is possible that I am missing something. Can someone help us please?

Answer 1

The textbook's given statement is wrong.

If $x$ is negative, then we can let $x = -a$ for any positive real number $a$. Then

$$\sqrt{x} \cdot \sqrt{x} = \sqrt{-a} \cdot \sqrt{-a} = i\sqrt{a} \cdot i\sqrt{a} = -\sqrt{a}^2 = -a = x.$$

Even with using imaginary numbers, which is a larger field than real numbers, the statement still doesn't make sense.

Answer 2

If you substitute -1 for x you get 1 as -x in the right side but in the left side you have i×i and that gives i² , that is -1 .
Therefore by counterexample method you can prove that the given statement is not valid for negative values of x .
I suggest here the reason is for fractional powers there exists more than one root.