# Math Help - Solving Inequality

1. ## Solving Inequality

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

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: $-\frac {2} {3} < x$

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 $[-\frac {2}{3}, 2]$

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

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

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: $-\frac {2} {3} < x$

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 $[-\frac {2}{3}, 2]$

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

Obviously , $x\neq 2$

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

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

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

Since $8x^2-10x+4$ is complex , then $2-x>0\Rightarrow x<2$

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

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: $-\frac {2} {3} < x$

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 $[-\frac {2}{3}, 2]$

somehow all my math programs give me the solution: -2/3 < x < 2
i cant see why that is
Clearly $x \neq 2$.

Case 1: $2 - x < 0$

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

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

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

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

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

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

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

Case 2: $2 - x > 0$.

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

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

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

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

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

This is true for all $x$.

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