1. ## Rational inequalities

3 different questions:

${\left({x}+{105}\right)}{\left({x}-{178}\right)}{\left({{x}}^{{2}}-{16}\right)}\gt{0}$

${\left({{x}}^{{2}}-{3}{x}\right)}{\left({x}+{7}\right)}\le-{1}{{x}}^{{2}}+{3}{x}$

$\frac{{{2}{{x}}^{{2}}-{29}}}{{{x}-{3}}}\ge{x}-{1}$

Put into interval notation.

question on the 1st one.. Should i factor out (X-4)(X+4) or leave it as is?
third one... do i subtract (x-1) from both sides... multiply by the LCD (x-3) ... that gives me (3x^2-4x-26)/(x-3) then using quadradic i get (2+-sqrt(82))/3..

2. Originally Posted by xsavethesporksx
3 different questions:

${\left({x}+{105}\right)}{\left({x}-{178}\right)}{\left({{x}}^{{2}}-{16}\right)}\gt{0}$

${\left({{x}}^{{2}}-{3}{x}\right)}{\left({x}+{7}\right)}\le-{1}{{x}}^{{2}}+{3}{x}$

$\frac{{{2}{{x}}^{{2}}-{29}}}{{{x}-{3}}}\ge{x}-{1}$

Put into interval notation.

question on the 1st one.. Should i factor out (X-4)(X+4) or leave it as is?
third one... do i subtract (x-1) from both sides... multiply by the LCD (x-3) ... that gives me (3x^2-4x-26)/(x-3) then using quadradic i get (2+-sqrt(82))/3..

Dear xsavethesporksx,

First step is to take all the terms into one side and factor the expresson. Then you could see for which values the inequality holds.

For example,

$(x+105)(x-178)(x^{2}-16)>{0}$

$(x+105)(x-178)(x-4)(x+4)>{0}$

$[x-(-105)](x-178)(x-4)[x-(-4)]>{0}$

Now draw a number line, you could consider the following situations one by one.

When, $x<-105\Rightarrow{(-)(-)(-)(-)}$ Therefore the result will be positive.

When, $-105 Therefore the result is negative.

When, $-4 Therefore the result is positive.

When, $4 Therefore the result is negative.

When, $x>178\Rightarrow{(+)(+)(+)(+)}$ Therefore the result is positive.

Therefore $-105>x~or~-4178$

$x\in{(-\infty,-105)\cup{(-4,4)}\cup{(178,\infty)}}$