summer101 wrote:

Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,

x> x^2

0>x(x-1)

Now 0>x and 0>(x-1)

for 0>x-1 ==> x <1, till here i am fine

but how do we get x>0 to make 0<x<1 ??

Till here you are fine x> x^2 so \(x^2-x<0\) we have to solve this, and to solve let me use an old trick.

Lets solve \(x^2-x=0,x(x-1)=0\) so x=1 or x=0. Now because the sign of x^2 is + and the operator is < we take the

INTERNAL values: \(0<x<1\).

Remember: to solve inequalities like this (x^2) treat them like equations (replace <,> with = ) then, once you have the results take a look at the sign of x^2 an the operator.

(<,-) or (>,+) take ESTERNAL values. (if they are the

"same")

(>,-) or (<,+) take INTERNAL values(like this case).

Let me know if it's clear now

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