Question

$\sf{\dfrac{x^2-5x+7}{-2x^2+3x+2}>0}$
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Solve this by wavy curve method.​

Answer 1

Given Question :-

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 5x + 7 }{ - {2x}^{2} + 3x + 2} > 0$

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

We know that,

If in a quadratic polynomial f(x) = ax² + bx + c,

$\sf \: a > 0 \: and \: {b}^{2} - 4ac < 0, \: then \: f(x) > 0$

Let consider, the expression

$\rm :\longmapsto\: {x}^{2} - 5x + 7$

we have,

$\rm :\longmapsto\:a = 1, \: b = - 5, \: c = 7$

$\rm :\implies\:a = 1 > 0 \: and \: {( - 5)}^{2} - 4 \times 7 = - 3 < 0$

$\bf\implies \: {x}^{2} - 5x + 7 > 0$

So,

Now Consider,

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 5x + 7 }{ - {2x}^{2} + 3x + 2} > 0$

$\rm :\longmapsto\:As \: {x}^{2} - 5x + 7 >0$

$\rm :\implies\: - {2x}^{2} + 3x + 2 > 0$

$\rm :\implies\: - ({2x}^{2} - 3x - 2) > 0$

$\rm :\implies\:{2x}^{2} - 3x - 2< 0$

$\rm :\longmapsto\:2 {x}^{2} - 4x + x - 2 < 0$

$\rm :\longmapsto\:2x(x - 2) + 1(x - 2) < 0$

$\rm :\longmapsto\:(2x + 1)(x - 2) < 0$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf interval & \sf sign \: of \: {2x}^{2} - 3x - 4 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < - \dfrac{1}{2} & \sf + ve \\ \\ \sf - \dfrac{1}{2} < x < 2 & \sf - \: ve \\ \\ \sf x > 2 & \sf + \: ve \end{array}} \\ \end{gathered}$

$\bf\implies \: - \dfrac{1}{2} < x < 2$

$\bf\implies \:x \in \: \bigg( - \dfrac{1}{2},2 \bigg)$