Equation involving a square root

[Ozma]

I have to solve [MATH]x+\sqrt{x+1}=0[/MATH]. My try is the following: since there is a square root, it must be [MATH]x \geq -1[/MATH].
The equation is equivalent to [MATH]x=-\sqrt{x+1}[/MATH].
Here is my first doubt: now the right hand side is always negative or zero, so I suppose I have to impose that the left hand side has that sign as well; so I must impose [MATH]x \leq 0[/MATH] too. Is this correct? If it is correct, I had to impose this because so I could go on with an equivalent equation or are there other reasons?
Going on I would square both sides so [MATH]x^2=x+1[/MATH] finding [MATH]x=\frac{1-\sqrt{5}}{2}[/MATH]; here is my second doubt: I know that [MATH]a=b \implies a^2=b^2[/MATH], but what is the meaning of that "implies" in this case? I mean, I am searching the [MATH]x[/MATH] that make [MATH]x+\sqrt{x+1}=0[/MATH] true, so why this method is correct to find [MATH]x[/MATH] if some steps like [MATH]x+\sqrt{x+1}=0\Leftrightarrow x=-\sqrt{x+1}[/MATH] have the equivalence "[MATH]\Leftrightarrow[/MATH]" while a step like [MATH]x=-\sqrt{x+1} \implies x^2=x+1[/MATH] has only the implication "[MATH]\implies[/MATH]"?
I think it's a logic thing, something like "The problem [MATH]x+\sqrt{x+1}=0[/MATH] has meaning for [MATH]x \geq -1[/MATH] and it is equivalent to the problem [MATH]x=-\sqrt{x+1}[/MATH] if [MATH]-1 \leq x \leq 0[/MATH], the problem [MATH]x=-\sqrt{x+1}[/MATH] causes (I used the word causes because I give this meaning to the "implication" sign) the fact that [MATH]x=\frac{1-\sqrt{5}}{2}[/MATH] and so this is the value you're searching for", but I'm not completely grasping why we're not interested in the other implication [MATH]\Leftarrow[/MATH], meaning "[MATH]x=\frac{1-\sqrt{5}}{2}[/MATH] implies [MATH]x+\sqrt{x+1}=0[/MATH]". So I'm basically confused by that "causes", because it is not "is equivalent"; why this solves the problem if it is not equivalent?
Hope this is not confusing, thanks to anyone who will help.

 

[Dr.Peterson]

When you take a step that only implies in one direction, you just have to check the result. You have found two solutions of the equation that results from squaring the original; the one that is not between -1 and 0 is an extraneous solution, and you just ignore it.

 

[pka]

I have to solve [MATH]x+\sqrt{x+1}=0[/MATH]. My try is the following: since there is a square root, it must be [MATH]x \geq -1[/MATH].
The equation is equivalent to [MATH]x=-\sqrt{x+1}[/MATH].
Here is my first doubt: now the right hand side is always negative or zero, so I suppose I have to impose that the left hand side has that sign as well; so I must impose [MATH]x \leq 0[/MATH] too. Is this correct? If it is correct, I had to impose this because so I could go on with an equivalent equation or are there other reasons?

Look at this link. Some times seeing is believing.

 

[Steven G]

That one way implication is really a two way implication if you include the restraints that you yourself imposed on x.

x = -sqrt(x+1) <==> x^2 = x+1 and -1<x<0. Those two statements are exactly the same.

 

[HallsofIvy]

Yes, from \(\displaystyle x+ sqrt{x+ 1}= 0\) you can deduce that, since a square root is never negative, in order to add to 0, must be a negative number. Subtracting \(\displaystyle \sqrt{x+ 1}\) from both sides, \(\displaystyle x= -\sqrt{x+ 1}\). Square both sides: \(\displaystyle x^2= x+ 1\) so \(\displaystyle x^2- x- 1= 0\). Use the quadratic formula to solve that: \(\displaystyle x= \frac{1\pm\sqrt{1- 4(1)(-1)}}{2}= \frac{1\pm\sqrt{5}}{2}\). Since \(\displaystyle \sqrt{5}> 1\) in order to have x positive we have to take the "+", not "-", square root: \(\displaystyle x= \frac{1+ \sqrt{5}}{2}\).