Can anyone help me with this quadratic inequality ? (1 Viewer)

[captain awesome]

(2-x)(x+1) >=0

Are you supposed to graph these on a parabola ?? :/

 

[Drongoski]

(2-x)(x+1) >=0

Are you supposed to graph these on a parabola ?? :/

If you expand the LHS this becomes: -x² + x + 2 > 0

The LHS is a quadratic in 'x'; if you sketch y = -x² + x + 2 you get an upside-down parabola intersecting the x-axis at x = -1 and at x = 2. When you look at the graph it becomes blindingly clear that the part of the graph is > 0 (i.e. positive; i.e. above the x-axis) for the interval -1 < x < 2

 

[SeCKSiiMiNh]

(2-x)(x+1) >=0

Are you supposed to graph these on a parabola ?? :/

my teacher taught me this method, which works fine imo.

you need to sketch the parabola.

x intercepts are 2 and -1 and since x^2 is positive, the parabola is concave upwards.

from the inequality, (2-x)(x+1) is greater than or equal 0, so you need to highlight the part of the graph ABOVE the x-axis. in the case that it is LESS than zero, you need the highlight the part of the graph BELOW the x-axis.
looking at the highlighted bit, you can note the you highlighted the part on the graph where x is less than or equal to -1 and the part where x is greater than or equal to 2. and those are your answers.

sorry if this sounds confusing.

but in short, sketch, shade and write your answer.

and it'll work for cubic equations as well