Question #67ae2

Question

Question #67ae2

1 Answer

Impact of this question

993 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-09T21:18:30

$] - \infty, - \frac{3}{2} [\cup] 0, 3 [$ $]-oo, -3/2[uu]0,3[$

Explanation:

$2 x^{3} < x^{2} + 3 x$ $2x^3 < x^2+3x$

$2 x^{3} - x^{2} - 3 x < 0$ $2x^3 - x^2-3x<0$

$x (2 x^{2} - x - 3) < 0$ $x(2x^2 - x-3)<0$

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2 \cdot (- 3)}}{4}$ $x=(-1+-sqrt(1-4*2*(-3)))/4$

$x = \frac{- 1 \pm \sqrt{25}}{4}$ $x=(-1+-sqrt(25))/4$

$x = \frac{- 1 \pm 5}{4}$ $x=(-1+-5)/4$

$x = - 1 or x = - \frac{3}{2}$ $x=-1 or x=-3/2$

The second-degree polynomial will be negative in $] - \frac{3}{2}, - 1 [$ $]-3/2,-1[$

#color(white)(...........)-oo.....-3/2......-1........0............# #xcolor(white)(...--)---------0++++++++polynomial color(white)(.)+++0color(white)(.)--0+++++gives color(white)(.)-----0color(white)(.)++0--0+++++#

$] - \infty, - \frac{3}{2} [\cup] 0, + \infty [$ $]-oo, -3/2[uu]0, +oo[$

Science

Math

Humanities

... and beyond

Question #67ae2

1 Answer

Explanation:

Related questions

Impact of this question