Can you solve the inequility, please

[yppetrov]

Can you solve the inequility, please

x² + 4x - 80 ≥ 0

[red_dog]

Solving the equation [tex]x^2+4x-80=0[/tex] we have [tex]x_1=-2-2\sqrt{21},x_2=-2+2\sqrt{21}[/tex].

Then, the solution of the inequality is [tex]x\in(-\infty,-2-2\sqrt{21}]\cup[-2+2\sqrt{21},\infty)[/tex]

[yppetrov]

Thank you!

what about
x³ + 4x - 80 ≥ 0

[red_dog]

The inequality can be written as [tex](x-4)(x^2+4x+20)\geq 0[/tex]
[tex]x^2+4x+20>0, \ \forall x\in\mathbf{R}[/tex].
So, [tex]x-4\geq 0\Rightarrow x\geq 4[/tex]

[Cock146]

This is to easy give me something to solve it...!!!

Thanks.



____________
watson

[Guest]

Yet another method:
[tex]x^2+ 4x- 80\ge 0[/tex]
add 80 to both sides
[tex]x^2+ 4x\ge 80[/tex]
Complete the square
[tex]x^2+ 4x+ 4\ge 84[/tex]
[tex](x+ 2)^2\ge 84= 4(21)[/tex]
[
The graph of [tex]y= (x+ 2)^2[/tex] is a parabola with vertex at (-2, 0) opening upward. It will be larger than 84 outside the interval from the negative root of [tex](x+ 2)^2= 84[/tex] to the positive root.

That is, [tex]x\le -2- 2\sqrt{21}[/tex] and [tex]x\ge -2+ 2\sqrt{21}[/tex].

[ammornil]

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