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This should be simple for me but I'm missing something. i start by dividing both sides by x which gives me $$x^2<2x+8$$ or I think it should, now my answer is $x \lt 10$ but that is not the right answer.

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so I should treat it like a polynomial. – Joshhw Feb 3 '14 at 19:11
It is a polynomial inequality so you can treat it like a polynomial. – John Habert Feb 3 '14 at 19:16

It is unclear which your question is. So here are answers to both.

Start by subtracting $2x$ from both sides, then we have x<2x+8 $$ -x<8 $$ Multiplying by $-1$ (then flipping the inequality sign), yields $$ x>-8 $$ or you could notice that $$ x<2x+8 $$ then subtracting $x$ from both sides $$ 0<x+8 $$ so $x>-8$.

EDIT. Your original question. Start with $x^2<2x+8$, then notice $$ \begin{align} x^2&<2x+8\\ x^2-2x-8&<0\\ (x+2)(x-4)&<0 \end{align} $$ For the left side to be negative, we want $x+2>0$ and $x-4<0$ or $x+2<0$ and $x-4>0$. The first means that $x>-2$ and $x<4$, that is $-2<x<4$. The second case says that $x<-2$ and $x>4$. But $x$ can't be bigger than $4$ and less than $-2$. So the solutions are those $x$ such that $-2<x<4$.

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The OP specified that the inequality was meant to be simply $x$ on the left hand side. – amWhy Feb 3 '14 at 19:10
Thank you, it's hard to answer questions when the OP flip-flops or is unclear. I shall amend my answer. – mathematics2x2life Feb 3 '14 at 19:12
the original problem was supposed to be $$x^2<2x+8$$ but I forgot to add it when I wrote it. I saw that I made the mistake and tried to fix it immediately, I think your answer helps me. I should treat it like a polynomial then? – Joshhw Feb 3 '14 at 19:14
It is a polynomial, so yes. Getting everything to one side and factoring tends to be the easiest way to do these. Then you just consider the $2$ cases. I have amended my answer to include both possible questions. – mathematics2x2life Feb 3 '14 at 19:15

$$x^2<2x+8\iff x^2-2x<8\iff x^2-2x+1<8+9 \\ \iff (x-1)^2 < 9 \iff |x-1|<3 \iff -3<x-1<3 \iff -2<x<4$$

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Be very careful about just dividing by $x$ when working with inequalities. For one thing, this can lead to division by $0,$ and for another, it can lead to silly conclusions. If we could simply divide by $x$ on both sides of $x^2<2+8,$ we "should" have concluded that $$x<\frac{2+8}x$$ (not $x<2+8$). This is nonsensical if $x=0,$ of course, but we also run into problems when $x$ is negative. For example, $(-1)^2=1<2+8,$ but $$-10=\frac{10}{-1}=\frac{2+8}{-1},$$ and we certainly can't say that $-1<-10.$

Instead, we could proceed as follows, using the difference of squares formula $a^2-b^2=(a+b)(a-b)$: $$x^2<10\\x^2-10<0\\x^2-\left(\sqrt{10}\right)^2<0\\\left(x+\sqrt{10}\right)\left(x-\sqrt{10}\right)<0$$ Now, the only way that a product of two real numbers can be negative is if one is negative and the other is positive. That means that $x+\sqrt{10}$ must be positive--so $x>-\sqrt{10}$--and $x-\sqrt{10}$ must be negative--so $x<\sqrt{10}.$ Thus, $-\sqrt{10}<x<\sqrt{10}.$

Edit: In response to your correction, we can take a similar approach to what we did above, but with a bit more work needed. $$x^2<2x+8\\x^2-2x<8\\x(x-2)<8\\(x-1+1)(x-1-1)<8\\(x-1)^2-1^2<8\\(x-1)^2-1<8\\(x-1)^2-9<0\\(x-1)^2-3^2<0\\(x-1+3)(x-1-3)<0\\(x+2)(x-4)<0$$ Hence, $x+2$ must be positive--so $x>-2$--and $x-4$ must be negative--so $x<4.$ Thus, $-2<x<4.$

Even in your corrected version, it is a bad idea to divide by $x.$ It "should" give us $x<\frac{2x+8}{x},$ which is once again nonsensical if $x=0$ and false when $x=-1$ (for example).

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$x^2-2x-8=0$, $x_{1,2}=\{4;-2\}$. Solution: $x \in (-2;4)$.

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We derive $(x-1)^2\le9\iff|x-1|\le3$. Now read that out loud: “The distance of $x$ to $1$ is at most $3$.”

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