Question

Solve the following inequality
(2x-7-5x^2)(x^2-5x+6)(x+1)>0

Answer 1

We're given to solve the inequality,

$\longrightarrow(2x-7-5x^2)(x^2-5x+6)(x+1)>0$

When we multiply both sides by -1 and make the term $2x-7-5x^2$ become $5x^2-2x+7$ the inequality symbol changes as,

$\longrightarrow(5x^2-2x+7)(x^2-5x+6)(x+1)<0\quad\quad\dots(1)$

For the term $5x^2-2x+7$ the discriminant is negative.

$\longrightarrow D=(-2)^2-4\cdot5\cdot7$

$\longrightarrow D=-136$

Since $a=5>0$ and $D=-136<0,$

$\longrightarrow5x^2-2x+7>0$

Therefore (1) implies,

$\longrightarrow(x^2-5x+6)(x+1)<0\quad\quad\dots(2)$

because a positive number is multipied with a negative number to get negative product.

Factorising $x^2-5x+6,$

$\longrightarrow x^2-5x+6=x^2-2x-3x+6$

$\longrightarrow x^2-5x+6=x(x-2)-3(x-2)$

$\longrightarrow x^2-5x+6=(x-2)(x-3)$

Then (2) becomes,

$\longrightarrow(x+1)(x-2)(x-3)<0$

Using wavy curve method,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(15,0)(15,0){3}{\circle{1.5}}\put(12,-5){ -1 }\put(29,-5){ 2 }\put(44,-5){ 3 }\multiput(0,-15)(45,15){2}{\qbezier(0,0)(7.5,7.5)(15,15)}\qbezier(15,0)(22.5,7.5)(30,0)\qbezier(30,0)(37.5,-7.5)(45,0)\end{picture}$

Hence the solution is,

$\longrightarrow\underline{\underline{x\in(-\infty,\ -1)\cup(2,\ 3)}}$