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

by **inequality** » Thu May 12, 2011 11:38 am

I need help to solve the inequality:
$(x^2-2x-8).(x-1)<0$

by **Math Tutor** » Fri May 13, 2011 9:12 am

Find the roots of the equation: $x^2-2x-8=0$

by **Guest** » Sun Jun 05, 2011 11:14 pm

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

by **Math Tutor** » Mon Jun 06, 2011 3:10 am

Nice solution, perfect explained!

by **Guest** » Mon Jun 06, 2011 12:22 pm

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

by **perfectmath** » Fri Jun 29, 2012 12:38 am

Nice solution!!!

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

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

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

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

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

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

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

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