find x again in equation

Question

I asked a similar question but I wanted to be sure understand. I have to find $x$ in the equation $$x^2=-2x-1$$ I go to left and get $$x^2+2x+1$$ Then I use a similar trick used in similar question and I get $$(x+1)^2$$ This I am not sure, but I think it is because $$(x+1)^2=(x+1)(x+1)=x(x+1)+1(x+1)=xx+x1+1x+11=x^2+x1+x1+1=x^2+x(1+1)+1=x^2+x2+1=x^2+2x+1$$ Here I used a trick in similar question to get the idea of $(x+1)^2=(x+1)(x+1)$. Then I expand like if $x$ is a number. Can I do this? Is x a number? If yes then I find $$(x+1)^2=0$$ Now I am not sure but this is a situation like finding the square root. I get $(x+1)=0$ because $\sqrt{0}=0$ as I asked in an other question. So $x+1=0$ and so $x=-1$. Hence I say that $$x=-1$$ is final answer. Is it good?

Answer 1

You ask a few questions here.

For the first, regarding why $x^2+2x+1=(x+1)^2$, your reasoning is correct. I think most people would have written it differently (the coefficient is usually written first, so you wouldn't have $x2$, and a coefficient of $1$ is usually omitted, so you would simply write $x$ instead of $x1$ or $1x$), but that isn't the focus of your question.

Then you ask "Is $x$ a number?". In this context, yes it is. The value of $x$ is currently unknown (meaning you don't know which number $x$ is), but that doesn't change the fact that it is a number.

So then you can take the square root of both sides of the equation $(x+1)^2=0$. However, if you are not comparing to zero, like if you had the equation $y^2=4$, you would need to consider both the positive and negative square roots. In this case, if $y^2=4$, then you could have $y=\sqrt{4}=2$ or $y=-\sqrt{4}=-2$, because both $2^2=4$ and $(-2)^2=4$. Since $-0=0$, though, you don't have to worry about that here.

Your final answer, then, is also correct. $x=-1$ is the solution to the equation.

Answer 2

Question: Can I expand variables like I would expand numbers? The answer is yes.

$x$ is a number. Also normally, when you take the square root of a square you put absolute value bars. However this is a special case (it equals 0) so the value inside the square must be also 0.

Answer 3

Your answer is correct. You found the zero of the function

(x+1)² = x² + x + 1 (x+1)(x+1) Now multiplay from left to right: xx + x*1 + 1*x + 1*1 and sum up: x² + 2x + 1