We’re given to solve the inequality, [tex]\longrightarrow(2x-7-5x^2)(x^2-5x+6)(x+1)>0[/tex] When we multiply both sides by -1 and make the term [tex]2x-7-5x^2[/tex] become [tex]5x^2-2x+7[/tex] the inequality symbol changes as, [tex]\longrightarrow(5x^2-2x+7)(x^2-5x+6)(x+1)<0\quad\quad\dots(1)[/tex] For the term [tex]5x^2-2x+7[/tex] the discriminant is negative. [tex]\longrightarrow D=(-2)^2-4\cdot5\cdot7[/tex] [tex]\longrightarrow D=-136[/tex] Since [tex]a=5>0[/tex] and [tex]D=-136<0,[/tex] [tex]\longrightarrow5x^2-2x+7>0[/tex] Therefore (1) implies, [tex]\longrightarrow(x^2-5x+6)(x+1)<0\quad\quad\dots(2)[/tex] because a positive number is multipied with a negative number to get negative product. Factorising [tex]x^2-5x+6,[/tex] [tex]\longrightarrow x^2-5x+6=x^2-2x-3x+6[/tex] [tex]\longrightarrow x^2-5x+6=x(x-2)-3(x-2)[/tex] [tex]\longrightarrow x^2-5x+6=(x-2)(x-3)[/tex] Then (2) becomes, [tex]\longrightarrow(x+1)(x-2)(x-3)<0[/tex] Using wavy curve method, [tex]\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}[/tex] Hence the solution is, [tex]\longrightarrow\underline{\underline{x\in(-\infty,\ -1)\cup(2,\ 3)}}[/tex] Reply
We’re given to solve the inequality,
[tex]\longrightarrow(2x-7-5x^2)(x^2-5x+6)(x+1)>0[/tex]
When we multiply both sides by -1 and make the term [tex]2x-7-5x^2[/tex] become [tex]5x^2-2x+7[/tex] the inequality symbol changes as,
[tex]\longrightarrow(5x^2-2x+7)(x^2-5x+6)(x+1)<0\quad\quad\dots(1)[/tex]
For the term [tex]5x^2-2x+7[/tex] the discriminant is negative.
[tex]\longrightarrow D=(-2)^2-4\cdot5\cdot7[/tex]
[tex]\longrightarrow D=-136[/tex]
Since [tex]a=5>0[/tex] and [tex]D=-136<0,[/tex]
[tex]\longrightarrow5x^2-2x+7>0[/tex]
Therefore (1) implies,
[tex]\longrightarrow(x^2-5x+6)(x+1)<0\quad\quad\dots(2)[/tex]
because a positive number is multipied with a negative number to get negative product.
Factorising [tex]x^2-5x+6,[/tex]
[tex]\longrightarrow x^2-5x+6=x^2-2x-3x+6[/tex]
[tex]\longrightarrow x^2-5x+6=x(x-2)-3(x-2)[/tex]
[tex]\longrightarrow x^2-5x+6=(x-2)(x-3)[/tex]
Then (2) becomes,
[tex]\longrightarrow(x+1)(x-2)(x-3)<0[/tex]
Using wavy curve method,
[tex]\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}[/tex]
Hence the solution is,
[tex]\longrightarrow\underline{\underline{x\in(-\infty,\ -1)\cup(2,\ 3)}}[/tex]