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.
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.
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