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Derivative from First Principle

Solve the fol...

Question

Solve the following inequality:
$\frac{30 x - 9}{x - 2} ⩾ 25 (x + 2)$

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Solution

$\frac{30 x - 9}{x - 2} \geq 25 (x + 2)$

$\frac{30 x - 9}{x - 2} - 25 (x + 2) \geq 0$

$= \frac{30 x - 9}{x - 2} - \frac{25 (x + 2) (x - 2)}{x - 2}$

$= \frac{30 x - 9 - 25 (x + 2) (x - 2)}{x - 2}$

$= \frac{- 25 x^{2} + 30 x + 91}{x - 2}$

$\frac{- 25 x^{2} + 30 x + 91}{x - 2} \geq 0$

$\frac{- (5 x + 7) (5 x - 13)}{x - 2} \geq 0$

$\frac{(5 x - 13) (5 x + 7)}{x - 2} \leq 0$

$5 x - 13 i s z e r o f o r : x = \frac{13}{5}$

$5 x + 7 i s z e r o f o r : x = - \frac{7}{5}$

$x - 2 i s z e r o f o r : x = 2$

$x \leq - \frac{7}{5} o r 2 < x \leq \frac{13}{5}$

$⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} S o l u t i o n : & x \leq - \frac{7}{5} o r 2 < x \leq \frac{13}{5} D e c i m a l : & x \leq - 1.4 o r 2 < x \leq 2.6 I n t e r v a l N o t a t i o n : & (- \infty, - \frac{7}{5}] \cup (2, \frac{13}{5}] \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

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