Inequality (x^2-2x-8).(x-1)<0

[inequality]

I need help to solve the inequality:
[tex](x^2-2x-8).(x-1)<0[/tex]

[Math Tutor]

Find the roots of the equation: [tex]x^2-2x-8=0[/tex]

[Guest]

[tex]x^{2}-2x-8=x^{2}-4x+2x-8=x(x-4)-2(x-4)=(x-2)(x-4)[/tex]
So [tex](x-1)(x-2)(x-4)<0[/tex].
If we have [tex]x > 4[/tex], then we take result with "+"
If we have [tex]2<x<4[/tex] , then we take result with "-"
If we have [tex]1<x<2[/tex], then we take result with "+"
If we have [tex]x<1[/tex], then we take result with "-"
But we must take results only with "-", so: [tex]x\in (-\infty ; 1)\cup (2;4)[/tex]

[Math Tutor]

Nice solution, perfect explained!

[Guest]

Math is too easy( I think in Europe and American school i met more interesting and diffucult tasks)

[perfectmath]

Nice solution!!!

[Xandra]

Thanks for solving this for me...

_______________________
Xandra

[Guest]

[tex]x^2- 2x- 8= (x- 4)(x+ 2)[/tex]
so [tex](x^2- 2x- 8 )(x- 1)= (x- 4)(x-(-2))(x- 1)\le 0[/tex]

For a product of three numbers to be negative either two of them are positive and one negative or all three are negative. We can order these numbers as -2< 1< 4. If x< -2 then all three factors are negative. If -2< x< 1 then x+ 2 positive while the other two factors are negative. If 1< x< 4 then both x+ 2 and x- 1 are positive while x- 4 is positive. If x> 4 then all three factors are positive.

[tex](x^2- 2x- 8 )(x- 1)< 0[/tex] for [tex]x< -2[/tex] and for [tex]1< x< 4[/tex].

[ammornil]

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