0

Also the solutions for x are apparently (according to another question I have posted)

$x=0$, $x=-1$, $x=-\frac{1}{3}$, $x=10$

But when graphing the line my self the solution are only $x=0$ and $x=-1$

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Thomas Winkworth
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  • Could you explain what you mean by "draw this line"? You've written an equation in one variable—what does that have to do with a line? – Matthew Leingang Feb 04 '15 at 12:02
  • I am honestly quite confused at this point as to what finding solutions really means, I always thought you assume y=0 if y isn't present in the equation? Does that make any sense? I am very confused right now to be honest... subtract the LHS make it equal to y then plot graph? :S I hope from this you can understand where I have gone wrong or perhaps I am going crazy – Thomas Winkworth Feb 04 '15 at 12:05
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    @MatthewLeingang Even though $x=-3$ is an equation written in one variable, it can certainly be represented by drawing a line in $\mathbb{R^2}$, but I agree that the question is poorly formulated. – Daniel W. Farlow Feb 04 '15 at 12:07
  • Is a graphical method the best way to solve this problem? – Matthew Leingang Feb 04 '15 at 12:10
  • @MatthewLeingang I think many of us still wonder what this problem is exactly. I simply wanted to point a potentially helpful observation. :) – Daniel W. Farlow Feb 04 '15 at 12:13
  • I guess a good question for me to ask at this point before I embarass myself anymore is "what is a solution?" – Thomas Winkworth Feb 04 '15 at 12:15

1 Answers1

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Normally to "solve" such equations, a common approach is to factorise. E.g. you can write the equation as $$x(x+1)\left(3(x^2+x+2)-16(2x+1) \right) = 0 \iff x(x+1)(1+3x)(x-10)=0$$

So we get $x = 0, -1, -\frac13, 10$ as solutions.

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