can anyone help me with this inequalities??

this is what i've done...

so....

or

and

roots rule

so the area will be at

or

but,

when i've try with x=2.5

then i'll be (it's absolutely wrong)

why????(Headbang)

regards,

pencil09

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- August 25th 2010, 07:25 AMpencil09inequalities
can anyone help me with this inequalities??

this is what i've done...

so....

or

and

roots rule

so the area will be at

or

but,

when i've try with x=2.5

then i'll be (it's absolutely wrong)

why????(Headbang)

regards,

pencil09 - August 25th 2010, 07:44 AMyeKciM
only solution is that

:D - August 25th 2010, 08:03 AMSudharaka
- August 25th 2010, 08:38 AMyeKciM
this way you can't be wrong :D

meaning you must look for the answer in sub intervals of interval

and you go easy :D

then calculate and see if the solutions are in interval if they are not then there is no solution in interval :D

then calculate and see if the solutions are in interval if they are not then there is no solution in interval :D

and of course for the

then calculate and see if the solutions are in interval if they are not then there is no solution in interval :D

after doing all of that, you will see which ones are the solutions :D - August 25th 2010, 08:53 AMumangarora
x^2 -7x + 18x has to be greater than 0..

as if its equated to 0, the roots are imaginary. - August 25th 2010, 09:04 AMumangarora
x-2 <= (x-4)^2..right??

Now, instead of adding/subtracting, u divide bothe sides by x-2

1<= (x-4)^2 / x- 2

this is easy to work with..

x cannot be less than 0(or 2).. that'd make the denominator negative, hence make the right side less than one.. which is not possible!

x=2 cannot be a sol. as that'd make the denominator 0.. which is not defined..

then x is between 4 and 2.. that'd make the right side a fraction less than 1.. again we reject this..

x=4 is not a sol. either as that'd make the right side 0.. hence we reject this.

now, try x greater than 4..IT WORKS :)

So, the ONLY SOLUTION IS x > 4 :)

PLZ THANK THIS POST ;) - August 25th 2010, 10:11 AMyeKciM
- August 25th 2010, 11:05 AMArchie Meade
Squaring can introduce "extraneous" solutions.

so, after you square, how will you tell the difference between those lines afterwards?

We can endeavour to solve using only algebra by not squaring as follows.

If this was zero, the roots would be...

from

Then the roots of the "quadratic"=0 are

leaving only one solution. - August 25th 2010, 11:27 AMyeKciM
by squaring inequalities, solutions of the inequalities will not change :D (#4) perhaps much work to be done, but is't simple work, nothing complicated, and for sure (because of simplicity) solutions are easy to get :D

- August 25th 2010, 12:58 PMpencil09
thanks for your answer.... (Clapping)

- August 25th 2010, 01:18 PMArchie Meade
You could simplify things for yourself by identifying whether or not you are accepting negative square roots of x-2.

In this case, it's not practical. hence....

in order that the value "under" the square root sign is not negative**and**the value of the square root is not negative.

Hence you rule out solutions <4.

Therefore, you can work with the quadratic formula that you created,

while disqualifying any solution less than 4. - August 25th 2010, 03:18 PMKrizalid
if there's no solution and that's clear, even if we restrict the radicand by setting then the only case of interest is when and then the radicand is well defined, and we can square, so we get which leads to and this is since we get which is the solution to our problem.

you may want to see this: http://www.mathhelpforum.com/math-he...es-132202.html