What is 4+1/(1-x)<= 9/x ?

The solution to 4+1/(1-x)<= 9/x is 0<x<= (7-sqrt(13))/4 \lor 1<x<= (7+sqrt(13))/4

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4+1/(1-x)<= 9/x

4 + \frac{1}{1 - x} \leq \frac{9}{x}

0 < x \leq \frac{7 - 13}{4} or 1 < x \leq \frac{7 + 13}{4}

Interval Notation

(0, \frac{7 - 13}{4}] \cup (1, \frac{7 + 13}{4}]

Decimal Notation

0 < x \leq 0.84861 \dots or 1 < x \leq 2.65138 \dots

Solution steps

4 + \frac{1}{1 - x} \leq \frac{9}{x}

Rewrite in standard form

\frac{- 4 x ^{2} + 14 x - 9}{x ( - x + 1 )} \leq 0

Factor

- 4 x^{2} + 14 x - 9 : - (x - \frac{7 + 13}{4}) (x - \frac{7 - 13}{4})

\frac{- ( x - \frac{7 + 13}{4} ) ( x - \frac{7 - 13}{4} )}{x ( - x + 1 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{7 + 13}{4} ) ( x - \frac{7 - 13}{4} ) ) ( - 1 )}{x ( - x + 1 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x - \frac{7 + 13}{4} ) ( x - \frac{7 - 13}{4} )}{x ( - x + 1 )} \geq 0

Identify the intervals

0 < x \leq \frac{7 - 13}{4} or 1 < x \leq \frac{7 + 13}{4}

4 a^{2} + 4 ab + b^{2}

(i)^{2}

(- x^{2} - 2 x + 8) = 0

\frac{1}{3} \cdot 18

\frac{3}{5} (\frac{x - 1}{3} + 1) + x = \frac{3}{4} (x - \frac{2}{3})

What is 4+1/(1-x)<= 9/x ?
The solution to 4+1/(1-x)<= 9/x is 0<x<= (7-sqrt(13))/4 \lor 1<x<= (7+sqrt(13))/4