Question

For what real value of x the following inequality holds:
(4x²)/[1 - √(1 + 2x)]² < 2x + 9 ?

Answer 1

(4x²)/[1 - √(1 + 2x)]² < 2x + 9
clearly the left side is always ≥ 0, (we are squaring the whole mess)
so 2x + 9 ≥ 0
x ≥ -9/2
furthermore, 1+2x > 0 or else we can't do √(1+2x)
x > -1/2
furthermore, 1 - √(1+x) ≠ 0
√(1+x) ≠ 1
1+x ≠ 1
x ≠ 0

then we have: x ≥ -9/2 AND x > -1/2 -----> x > -1/2
<b>so x > -1/2 , x ≠ 0</b>

Answer 2

Thank you sir reiny......

Answer 3

I just notice a typo

in:
furthermore, 1 - √(1+x) ≠ 0
it should say <b> furthermore, 1 - √(1+2x) ≠ 0</b>
√(1+2x) ≠ 1
1+2x ≠ 1
x ≠ 0 <----- does not change the solution

Answer 4

I see

Answer 5

What if x=4? Then you have
(4*16)/(1-3)^2 < 2*4+9
16 < 17
So, what's missing?
(4x²)/[1 - √(1 + 2x)]² < 2x + 9
4x^2 < (2x+9)(1-2√(1+2x)+1+2x)
2x^2 < (2x+9)(x+1-√(1+2x))
(2x+9)√(1+2x) < 11x+9
(2x+9)^2 (1+2x) < (11x+9)^2
8x^3 + 76x^2 + 198x + 81 < 121x^2 + 198x + 81
8x^3 - 45x^2 < 0
x^2(8x-45) < 0
x < 45/8
The original equation clearly requires that x > -1, and x≠0
So, x is in [-1,0)U(0,45/8)

Answer 6

Nice solution oobleck.
I graphed it in Wolfram and got 45/8 as I was working it,
but could not reconcile it with my solution.
https://www.wolframalpha.com/input/?i=solve++%284x%C2%B2%29%2F%5B1+-+%E2%88%9A%281+%2B+2x%29%5D%C2%B2++%3D++2x+%2B+9

Can you see a flaw in what I did?

Answer 7

Okay thank you