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????
regards,
pencil09
this way you can't be wrong
meaning you must look for the answer in sub intervals of interval
and you go easy
then calculate and see if the solutions are in interval if they are not then there is no solution in interval
then calculate and see if the solutions are in interval if they are not then there is no solution in interval
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
after doing all of that, you will see which ones are the solutions
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
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.
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.
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