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Mathematics 18 Online
OpenStudy (anonymous):

Modulus

mathslover (mathslover):

Cool! :P

mathslover (mathslover):

What is ur question?

OpenStudy (anonymous):

\[\left| \left| x \right|-1 \right| < \left| 1-x \right|\]

OpenStudy (rsadhvika):

square both sides

mathslover (mathslover):

Always try to remove that modulus sing there. It makes the work easier.

mathslover (mathslover):

(|x| - 1)^2 < (1-x)^2 Now, solve !

OpenStudy (anonymous):

\[\left| x \right|^{2} -2\left| x \right| +1 = x ^{2}-3x +1\]

mathslover (mathslover):

LHS is right... RHS - check again!

OpenStudy (anonymous):

-2x typo

mathslover (mathslover):

Fine. Now, see |x|^2 = x^2, right?

OpenStudy (anonymous):

\[-2\left| x \right| = -2x\]

mathslover (mathslover):

Good. You have |x| = x , right?

OpenStudy (anonymous):

yeah

mathslover (mathslover):

o.O don't forget the inequality " < "

mathslover (mathslover):

\(-2 |x| < -2x\) Divide -2 both sides : \(|x| > x\)

OpenStudy (anonymous):

-1

OpenStudy (anonymous):

-1 is the answer

OpenStudy (anonymous):

I am not able to delete my comments are you

mathslover (mathslover):

Yep.. I can delete my comments.

mathslover (mathslover):

The solution for this is x < 0 only..

mathslover (mathslover):

You can not say that the answer is -1 or something... put -2 at the place of x |2 - 1| < |1+2| 1 < 3 which is true so, any value for x which is less than 0 will satisfy the condition.

OpenStudy (anonymous):

wait i am reading my internet wasn't working

mathslover (mathslover):

well, we have : |x| > x x can be anything but less than zero.

OpenStudy (anonymous):

yes all negative numbers

OpenStudy (anonymous):

yeah , we cannot say exactly -1

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