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?
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Your answer is correct. – recmath Dec 12 '14 at 21:30
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You should remember to mention that $x^2 + 2x + 1 = 0$ in your second equation. If you do not mention that it equals zero, your instructor may deduct marks. – Dunka Dec 12 '14 at 21:52
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Thank you @Dunka for the advice. I will remember it. Maybe because of you I will have better mark! I love this site. Please look at my other question. – Jacob Dec 12 '14 at 21:55
3 Answers
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.
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I very like your answer. Thank you. I also like turntable. My dad as one and we like to listen to music with it. We listen classical music like Bach and Beethoven. What do you listen on your turntable? – Jacob Dec 12 '14 at 21:43
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.
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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
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