# Solving inequalities

• May 24th 2011, 01:05 PM
Arron
Solving inequalities
Hi! Can anyone help? I need to solve the following inequalitiy

$\frac{{3x}^{2 }-3x-5 }{{x}^{2 }-2x-3 } >3$

I rearrange the inequalitiy first into

$\frac{{3x}^{2 }-3x-5 }{{x}^{2}-2x-3 }-3>0$

I the need to use the mulitiplication rule but am not sure what to multiply by...
• May 24th 2011, 01:20 PM
girdav
Write $\frac{3x^2-3x-5}{x^2-2x-3}-3 =\frac{3x^2-3x-5-3(x^2-2x-3)}{x^2-2x-3}$ and factorize the denominator.
• May 24th 2011, 01:30 PM
Deveno
note that 3x^2 - 3x - 5 = 3(x^2 - 2x - 3) + 3x + 4.

so (3x^2 - 3x - 5)/(x^2 - 2x - 3) = 3 + (3x + 4)/(x^2 - 2x - 3).

clearly, this will be > 3 iff (3x + 4)/(x^2 - 2x - 3) > 0.

now (3x + 4)/(x^2 - 2x - 3) = (3x + 4)/[(x - 3)(x + 1)].

for this to be positive, the numerator and the denominator must both have the same sign.

suppose 3x + 4 > 0, that is x > -4/3.

if -4/3 < x < -1, then x+1 is negative, and x-3 is also negative, so the denominator and numerator are both positive.

if -1 < x < 3, then x+1 is positive, and x-3 is negative, so the denominator is negative, so these values won't work.

if 3 < x, then x+1 is positive, and x-3 is positive, so the denominator and the numerator are both positive.

now, suppose x < -4/3. then x+1 and x-3 are always negative, so the denominator is positive, so these values won't work.

so the only values of x for which (3x^2 - 3x - 5)/(x^2 - 2x - 3) > 3 are in the intervals (-4/3,-1), and (3,∞).
• May 24th 2011, 02:32 PM
bigwave
also this can be derived to $(x-3)(x+1)(3x+4)<0$

after test points you get

$x>3$ and $-\frac{4}{3 } < x < -1$