Math, asked by niamthao2502, 5 hours ago

Solve for x:x2-x-30/x-1_>0

Answers

Answered by mathdude500
3

\large\underline{\sf{Solution-}}

Giving inequality is

\rm :\longmapsto\:\dfrac{ {x}^{2} - x - 30 }{x - 1}  \geqslant 0 \:  \: and \: x \ne \: 1

\rm :\longmapsto\:\dfrac{ {x}^{2} -6 x  + 5x- 30 }{x - 1}  \geqslant 0 \:  \: and \: x \ne \: 1

\rm :\longmapsto\:\dfrac{ x(x - 6) + 5(x - 6)}{x - 1}  \geqslant 0 \:  \: and \: x \ne \: 1

\rm :\longmapsto\:\dfrac{ (x - 6)(x + 5)}{x - 1}  \geqslant 0 \:  \: and \: x \ne \: 1

Now, critical points are - 5, 1 and 6

\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( -  \infty , - 5) & \sf   < 0 \\ \\ \sf [ - 5,1) & \sf  \geqslant 0 \\ \\ \sf (1,6) & \sf   < 0\\ \\ \sf [6, \infty ) & \sf   \geqslant 0 \end{array}} \\ \end{gathered}

So,

\bf\implies \:x \:  \in \: [ - 5,1) \:  \cup \: [6, \infty )

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Additional Information

\boxed{\tt{ x > y \:  \: \rm\implies \: \:  - x <  - y \: }}

\boxed{\tt{ x  \geqslant  y \:  \: \rm\implies \: \:  - x  \leqslant   - y \: }}

\boxed{\tt{  |x|  < y \:  \: \rm\implies \: \:  - y < x < y \: }}

\boxed{\tt{  |x|  \leqslant  y \:  \: \rm\implies \: \:  - y  \leqslant  x  \leqslant y \: }}

\boxed{\tt{  |x| > y \:  \: \rm\implies \: \: x <  - y \:  \: or \:  \: x > y \: }}

\boxed{\tt{  |x| \geqslant y \:  \: \rm\implies \: \: x \leqslant   - y \:  \: or \:  \: x  \geqslant  y \: }}

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