Solving Inequality

• Sep 23rd 2009, 07:09 AM
coobe
Solving Inequality
$\displaystyle \frac {-4x^2} {2-x} < 2-4x$

im not sure about my result.

as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

which leads me to the first set of solutions: x > 2

am i right here ?

the 2nd case would give me the same just with flipped inequality sign

so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is :(
• Sep 23rd 2009, 07:28 AM
Quote:

Originally Posted by coobe
$\displaystyle \frac {-4x^2} {2-x} < 2-4x$

im not sure about my result.

as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

which leads me to the first set of solutions: x > 2

am i right here ?

the 2nd case would give me the same just with flipped inequality sign

so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is :(

HI

Obviously , $\displaystyle x\neq 2$

$\displaystyle -\frac{4x^2}{2-x}<2-4x$

$\displaystyle 0<2-4x+\frac{4x^2}{2-x}$

$\displaystyle \frac{8x^2-10x+4}{2-x}>0$

Since $\displaystyle 8x^2-10x+4$ is complex , then $\displaystyle 2-x>0\Rightarrow x<2$
• Sep 23rd 2009, 07:39 AM
Prove It
Quote:

Originally Posted by coobe
$\displaystyle \frac {-4x^2} {2-x} < 2-4x$

im not sure about my result.

as far i can see i must look at 2 cases. 1st: x > 2 and 2nd x < 2

the first would give me the solution: $\displaystyle -\frac {2} {3} < x$

which leads me to the first set of solutions: x > 2

am i right here ?

the 2nd case would give me the same just with flipped inequality sign

so the overall solution i got is: all real numbers except the intervall $\displaystyle [-\frac {2}{3}, 2]$

somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is :(

Clearly $\displaystyle x \neq 2$.

Case 1: $\displaystyle 2 - x < 0$

$\displaystyle \frac{-4x^2}{2 - x} < 2 - 4x$

$\displaystyle -4x^2 > (2 - 4x)(2 - x)$

$\displaystyle -4x^2 > 4 - 10x + 4x^2$

$\displaystyle 8x^2 - 10x + 4 < 0$

$\displaystyle x^2 - \frac{5}{4}x + \frac{1}{2} < 0$

$\displaystyle x^2 - \frac{5}{4}x + \left(-\frac{5}{8}\right)^2 - \left(-\frac{5}{8}\right)^2 + \frac{1}{2} < 0$

$\displaystyle \left(x - \frac{5}{8}\right)^2 + \frac{7}{64} < 0$

Case 2: $\displaystyle 2 - x > 0$.

$\displaystyle \frac{-4x^2}{2 - x} < 2 - 4x$

$\displaystyle -4x^2 < (2 - 4x)(2 - x)$

$\displaystyle -4x^2 < 4 - 10x + 4x^2$

$\displaystyle 8x^2 - 10x + 4 > 0$

$\displaystyle \left(x - \frac{5}{8}\right)^2 + \frac{7}{64} > 0$

This is true for all $\displaystyle x$.

Thus, the solution to the inequality is $\displaystyle 2 - x > 0$, or $\displaystyle x < 2$.