1. ## inequality 2

I couldn't solve the following questions.
1)|x^2+x-5|>2X+1
2)|X-2|-|X+1|-X+2<0

2. ## Re: inequality 2

In the first, square both sides, then solve the resulting polynomial.

3. ## Re: inequality 2

You don't have to square both sides.
x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.

4. ## Re: inequality 2

Originally Posted by kastamonu
You don't have to square both sides.
x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.
true you don't have to, but doing so gives you the same answer and is easier to solve than the absolute value equations.

5. ## Re: inequality 2

Originally Posted by kastamonu
You don't have to square both sides.
x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.
looking at the first one and applying Prove It's method.

$|x^2+x-5|> 2x+1 \Rightarrow (x^2+x-5)^2> (2x+1)^2$

$x^4+2 x^3-9 x^2-10 x+25>4 x^2+4 x+1$

$x^4+2 x^3-13 x^2-14 x+24>0$

whittling this down by checking for factors of the form $(x\pm r)$ where r divides 24 we eventually reduce this to

$(x-3) (x-1) (x+2) (x+4)>0$

checking signs from $-\infty \to \infty$ we get

$(\infty, -4) \rightarrow +$

$(-4, -2) \rightarrow -$

$(-2, 1) \rightarrow +$

$(1, 3) \rightarrow -$

$(3,\infty) \rightarrow +$

and thus our solution is

$(\infty, -4) \cup (-2, 1) \cup (3, \infty)$

Solving that is much more methodical than trying to solve the absolute value equation directly.

Apply this to the second problem and see if you can work it out.

6. ## Re: inequality 2

Second one can be solved by this way:
1)If -1<x<2 then x>1
2)if x<-1 then
-x+2+x+1-x+2<0
x>5
According to my book answer is x>1.Then it is correct.