Math Help - inequalities

1. 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)

why????

regards,
pencil09

2. only solution is that

$6\le x$

3. 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)

why????

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...

4. 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)

why????

regards,
pencil09
this way you can't be wrong

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

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

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

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

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

and you go easy

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

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

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

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

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

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

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 $x \in (4,+\infty)$

after doing all of that, you will see which ones are the solutions

5. x^2 -7x + 18x has to be greater than 0..
as if its equated to 0, the roots are imaginary.

6. 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

7. 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 $x \in \mathbb{R} \;\;\; , 6\le x$

8. 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)

why????

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.

9. by squaring inequalities, solutions of the inequalities will not change (#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

11. 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.

12. 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