Question about the square root of a non-negative quadratic function.

Question

I encountered this question on a test:

For which values of $x$ does this apply $\sqrt{x^2 + 2x + 1} = x + 1$

Since the function $f(x) = x^2 + 2x +1$ never goes below $0$, I picked the answer "for all real numbers", this however was incorrect. The correct answer was $x \ge -1$ and I don't understand why.

I realize that $x^2 + 2x +1 = (x+1)^2$, but this shouldn't matter, should it?

Answer 1

You have$$\sqrt{x^2+2x+1}=\sqrt{(x+1)^2}=|x+1|.$$And$$|x+1|=x+1\iff x+1\geqslant0\iff x\geqslant-1.$$

Answer 2

Well, $(x+1)\geq0\iff x\ge-1$. So whenever $x\geq -1$, $\sqrt{x^2+2x+1}=x+1$. Your'e right that the square root returns positive values, but you should notice that the positive values are of $x+1$ and not $x$. Another way to see it is to replace $x+1$ with $y$. Now the question becomes - when does $\sqrt{y^2}=y$. This is ofcourse true $\iff$ $y\geq 0\iff x+1\geq0$

Answer 3

You have $$ \sqrt{x^2+2x+1} = \sqrt{(x+1)^2} $$

We know that $a^2>0$. Using this in above equation.

We get $$ x+1>0 $$ (How?)

So we get $x \geqslant 1$