Question

Represent the solution on number line

$\frac{ {8x}^{2} + 16x - 51}{(2x - 3)(x + 4)} < 3$
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Answer 1

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

Given inequality is

$\rm \: \dfrac{ {8x}^{2} + 16x - 51}{(2x - 3)(x + 4)} < 3$

can be rewritten as

$\rm \: \dfrac{ {8x}^{2} + 16x - 51}{(2x - 3)(x + 4)} - 3 < 0$

$\rm \: \dfrac{ {8x}^{2} + 16x - 51 - 3(2x - 3)(x + 4)}{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ {8x}^{2} + 16x - 51 - 3(2 {x}^{2} + 8x - 3x - 12)}{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ {8x}^{2} + 16x - 51 - 3(2 {x}^{2} + 5x - 12)}{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ {8x}^{2} + 16x - 51 - 6 {x}^{2} - 15x + 36}{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ {2x}^{2} +x -15 }{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ {2x}^{2} +6x - 5x -15 }{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{ 2x(x + 3) - 5(x + 3)}{(2x - 3)(x + 4)} < 0$

$\rm \: \dfrac{(x + 3)(2x - 5)}{(2x - 3)(x + 4)} < 0$

So, breaking points are

$\rm \: x = - 4, \: - 3, \: \dfrac{3}{2}, \: \dfrac{5}{2}$

So, sign of expression in intervals are as

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty , - 4) & \sf + ve \\ \\ \sf ( - 4, - 3) & \sf - ve \\ \\ \sf \bigg( - 3,\dfrac{3}{2} \bigg) & \sf + ve\\ \\ \sf \bigg(\dfrac{3}{2} ,\dfrac{5}{2} \bigg) & \sf - ve\\ \\ \sf \bigg(\dfrac{5}{2} , \infty \bigg) & \sf + ve \end{array}} \\ \end{gathered}$

So,

$\rm\implies \:\boxed{ \rm{ \rm  \:x \in \: ( - 4, - 3) \: \cup \: \bigg(\dfrac{3}{2}, \frac{5}{2} \bigg) \: \: }}$

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Additional Information :-

If a and b are real positive numbers such that a < b, then

$\boxed{\rm{ \: (x - a)(x - b) < 0 \: \: \rm\implies \:a < x < b \: }}$

$\boxed{\rm{ \: (x - a)(x - b) \leqslant 0 \: \: \rm\implies \:a \leqslant x \leqslant b \: }}$

$\boxed{\rm{ \: (x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x \leqslant a \: \: or \: x \geqslant b \: }}$

$\boxed{\rm{ \: (x - a)(x - b) > 0 \: \: \rm\implies \:x < a \: \: or \: x > b \: }}$