Set of solutions to $\sqrt{x^2}=-x$

Question

The question is:

The set of all real numbers x such that $$\sqrt{x^2}=-x$$ consists of:

A: zero only

B: nonpositive real numbers only

C: positive real numbers only

D: all real numbers

E: no real numbers

I chose D because the root of any number gives a $+$ and $-$, so if $x$ is a positive real number eg. $2$ then $\sqrt{4^2}= -2$ and if $x$ is a negative real number eg. $-2$ then $\sqrt{4^2} = 2$

But the answer is B, can someone explain this to me? I'm presuming that in maths when they have an equation with $\sqrt{\cdot}$ they mean $+$ unless there is a $-$ or plus/minus sign. Can someone confirm this?

Answer 1

The square root of a nonnegative real number $x$ is, by definition, the positive real number $a$ such that $a^2 = x$. As a consequence, you have the property $$\sqrt{x^2} = |x|$$ which reduces your question to: what real numbers satisfy $$|x| = -x$$ If you know how the absolute value works (or: check its definition), this shouldn't be hard anymore.

Answer 2

Let's use your example.

Putting $x=2$ in the given function we get $\sqrt {2^2}=-2$.

But we know that square roots always produce positive numbers.

Hence $\sqrt {2^2}$ should be equal to $2$. Therefore if $x$ is positive, then the given equation won't work.

Now, putting $x=-2$ in the given equation we get $\sqrt {(-2)^2}=-(-2)=2.$ Which is true since $\sqrt {(-2)^2}=\sqrt {2^2}=\sqrt 4=2$

So in other words, the given equation holds for all negative real numbers and Zero.

Hence B is the correct option.