1. ## Solve this inequality

$|3x+1|>\sqrt{x+1}$

My working :

There will be 2 cases here .

Case 1 : $3x+1>\sqrt{x+1}$

$9x^2+6x+1>x+1$

$x(9x+5)>0$

Case 2 : $3x+1<-\sqrt{x+1}$

$(3x+1)^2>x+1$

$x(9x+5)>0$ , Isn't this the same as case 1 ..

I think i made a mistake at case 2 . Can someone check for me .. Thanks .

2. second case is irrelevant since both sides are positive, so you are free to square them and solve the quadratic inequality, provided of course that we require $x\ge-1.$

3. Originally Posted by Krizalid
second case is irrelevant since both sides are positive, so you are free to square them and solve the quadratic inequality, provided of course that we require $x\ge-1.$
Thanks Krizalid .

So i will only need to solve $x(9x+5)>0$ to get my solution right ?

If so , my answer is $(-\frac{5}{9},\infty)$

But the answer given is ${x:x>0 or -1 .. Puzzled again

4. that's because you need to restrict the square root.

solution set for the quadratic inequality is $\left] -\infty ,-\frac{5}{9} \right[\cup \left] 0,\infty \right[.$ but this is not the full solution, 'cause on the other hand we require $x\ge-1,$ thus the solution set it's actually $\left] -\infty ,-\frac{5}{9} \right[\cup \left] 0,\infty \right[\cap \left[ -1,\infty \right[=\left[ -1,-\frac{5}{9} \right[\cup \left] 0,\infty \right[,$ which is our full solution set.

5. Another resolution

6. you did the same i did.