# inequalities

• Aug 25th 2010, 08:25 AM
pencil09
inequalities
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

this is what i've done...
$(\sqrt{x-2})^2 \leq (x-4)^2$
$x-2\leq x^2-8x+16$
so....
$x\leq 3$or $x\geq6$

and
roots rule

$x-2\geq0$
$x\geq2$

so the area will be at
$2\leqx\leq3$ or $x\geq6$

but,
when i've try with x=2.5
then i'll be $\sqrt{0.5} \leq -1.5$ (it's absolutely wrong)

regards,
pencil09
• Aug 25th 2010, 08:44 AM
yeKciM
only solution is that

$6\le x$

:D
• Aug 25th 2010, 09:03 AM
Sudharaka
Quote:

Originally Posted by pencil09
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

this is what i've done...
$(\sqrt{x-2})^2 \leq (x-4)^2$
$x-2\leq x^2-8x+16$
so....
$x\leq 3$or $x\geq6$

and
roots rule

$x-2\geq0$
$x\geq2$

so the area will be at
$2\leqx\leq3$ or $x\geq6$

but,
when i've try with x=2.5
then i'll be $\sqrt{0.5} \leq -1.5$ (it's absolutely wrong)

regards,
pencil09

Dear pencil09,

It is incorrect to say that, $x\geq{2}$ because,

$0\leq{\sqrt{x-2}}\leq{x-4}\Rightarrow{x\geq{2}~and~x\geq{4}}\Rightarrow{x\ geq{4}}$

Hope you could continue from here...
• Aug 25th 2010, 09:38 AM
yeKciM
Quote:

Originally Posted by pencil09
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

this is what i've done...
$(\sqrt{x-2})^2 \leq (x-4)^2$
$x-2\leq x^2-8x+16$
so....
$x\leq 3$or $x\geq6$

and
roots rule

$x-2\geq0$
$x\geq2$

so the area will be at
$2\leqx\leq3$ or $x\geq6$

but,
when i've try with x=2.5
then i'll be $\sqrt{0.5} \leq -1.5$ (it's absolutely wrong)

regards,
pencil09

this way you can't be wrong :D

$\sqrt{x-2} \le x-4$

$(x-2)\le (x-4)^2$

$x-2\neq 0 \Rightarrow x\neq 2$

$x-4\neq 0 \Rightarrow x\neq 4$

meaning you must look for the answer in sub intervals $(-\infty , 2) , (2,4) , (4, +\infty)$ of interval $(-\infty , + \infty)$

and you go easy :D

$x\in (-\infty , 2 ) \Rightarrow (x-2) < 0 \wedge (x-4)^2 >0$

$-(x-2) \le (x-4)^2$

then calculate and see if the solutions are in interval $(-\infty ,2)$ if they are not then there is no solution in interval $x\in (-\infty ,2)$ :D

$x\in (2 , 4 ) \Rightarrow (x-2) > 0 \wedge (x-4)^2 >0$

$(x-2) \le (x-4)^2$

then calculate and see if the solutions are in interval $(2,4)$ if they are not then there is no solution in interval $x \in (2,4)$ :D

and of course for the

$x\in (4 , +\infty ) \Rightarrow (x-2) > 0 \wedge (x-4)^2 >0$

$(x-2) \le (x-4)^2$

then calculate and see if the solutions are in interval $(4,+\infty)$ if they are not then there is no solution in interval $x \in (4,+\infty)$ :D

after doing all of that, you will see which ones are the solutions :D
• Aug 25th 2010, 09:53 AM
umangarora
x^2 -7x + 18x has to be greater than 0..
as if its equated to 0, the roots are imaginary.
• Aug 25th 2010, 10:04 AM
umangarora
x-2 <= (x-4)^2..right??

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 ;)
• Aug 25th 2010, 11:11 AM
yeKciM
Quote:

Originally Posted by umangarora

So, the ONLY SOLUTION IS x > 4 :)

no it's not ...

x=5

$\sqrt{3} \le 1$

you say this is true ???

so it's for every $x \in \mathbb{R} \;\;\; , 6\le x$
• Aug 25th 2010, 12:05 PM
Quote:

Originally Posted by pencil09
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

this is what i've done...

$(\sqrt{x-2})^2 \leq (x-4)^2$

$x-2\leq x^2-8x+16$

so....

$x\leq 3$or $x\geq6$

which is true for the quadratic equation, but are both of these true for the inequality ?

and
roots rule

$x-2\geq0$

yes, so that you are taking the square root of a positive value

$x\geq2$

so the area will be at

$2\leqx\leq3$ or $x\geq6$

but,
when i've try with x=2.5

then i'll be $\sqrt{0.5} \leq -1.5$ (it's absolutely wrong)

regards,
pencil09

Squaring can introduce "extraneous" solutions.

$(x-4)^2=(4-x)^2$

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.

$x-4\ \ge\ \sqrt{x-2}$

$(x-2)-\sqrt{x-2}-2\ \ge\ 0$

$\left(\sqrt{x-2}\right)^2-\sqrt{x-2}-2\ \ge\ 0$

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

$\sqrt{x-2}=\displaystyle\frac{1\pm\sqrt{1-4(-2)}}{2}$

from

$ay^2+by+c=0\Rightarrow\ y=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Then the roots of the "quadratic"=0 are

$\sqrt{x-2}=\displaystyle\frac{1\pm3}{2}=2,\;-1$

$\sqrt{x-2}\text{\footnotesize\;\;must be positive}\Rightarrow\ \sqrt{x-2}\ \ne\ -1$

leaving only one solution.
• Aug 25th 2010, 12:27 PM
yeKciM
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
• Aug 25th 2010, 01:58 PM
pencil09
• Aug 25th 2010, 02:18 PM
Quote:

Originally Posted by pencil09
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

this is what i've done...

$(\sqrt{x-2})^2 \leq (x-4)^2$

$x-2\leq x^2-8x+16$

so....

$x\leq 3\text{\footnotesize\;\;or\;\;}x\geq6$

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....

$x-2\ \ge\ 0\text{\footnotesize\;\;and\;\;}x-4\ \ge\ 0$

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.
• Aug 25th 2010, 04:18 PM
Krizalid
Quote:

Originally Posted by pencil09
can anyone help me with this inequalities??

$\sqrt{x-2}\leq x-4$

if $x<4,$ there's no solution and that's clear, even if we restrict the radicand by setting $x\ge2,$ then the only case of interest is when $x>4,$ and then the radicand is well defined, and we can square, so we get $x-2\le x^2-8x+16$ which leads to $x^2-9x+18\ge0$ and this is $(x-3)(x-6)\ge0,$ since $x>4$ we get $x\ge6,$ which is the solution to our problem.

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