Question

1) find the solution set of the following inequality
$2x^2 +5x + 2 > 0$
$(2 + x)(x - 5)(x + 1) > 0$
$\frac{2x - 4}{x + 3} > \frac{x + 2}{2x + 6}$

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Answer 1

1) Let us find the solution to the inequality,

$\longrightarrow2x^2+5x+2>0$

Factorising LHS by splitting middle term,

$\longrightarrow2x^2+4x+x+2>0$

$\longrightarrow2x(x+2)+x+2>0$

$\longrightarrow(x+2)(2x+1)>0$

$\Longrightarrow\underline{\underline{x\in\bigg(-\infty,\ -2\bigg)\cup\left(-\dfrac{1}{2},\ \infty\right)}}$

This is the solution to the inequality.

2) We make use of wavy curve method to find the solution to the inequality,

$\longrightarrow(x+2)(x-5)(x+1)>0$

The wavy curve is as shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(15,0)(15,0){3}{\circle{1.5}}\qbezier(45,0)(52.5,7.5)(60,15)\qbezier(30,0)(37.5,-7.5)(45,0)\qbezier(15,0)(22.5,7.5)(30,0)\qbezier(0,-15)(7.5,-7.5)(15,0)\small\put(13,-4){-2}\put(28,-4){-1}\put(44,-4){5}\end{picture}$

Hence the solution to the inequality is,

$\longrightarrow\underline{\underline{x\in(-2,\ -1)\cup(5,\ \infty)}}$

3) Given,

$\longrightarrow\dfrac{2x-4}{x+3}>\dfrac{x+2}{2x+6}$

Multiplying both sides by 2 we get,

$\longrightarrow\dfrac{4x-8}{x+3}>\dfrac{x+2}{x+3}$

Taking RHS to LHS,

$\longrightarrow\dfrac{4x-8}{x+3}-\dfrac{x+2}{x+3}>0$

$\longrightarrow\dfrac{4x-8-x-2}{x+3}>0$

$\longrightarrow\dfrac{3x-10}{x+3}>0$

$\Longrightarrow\underline{\underline{x\in\bigg(-\infty,\ -3\bigg)\cup\left(\dfrac{10}{3},\ \infty\right)}}$

This is the solution to the inequality.