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Math Help - inequalities

  1. #1
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    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 3or 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
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  2. #2
    Senior Member yeKciM's Avatar
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    only solution is that

     6\le x

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  3. #3
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    Quote Originally Posted by pencil09 View Post
    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 3or 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...
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by pencil09 View Post
    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 3or 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
    Last edited by yeKciM; August 25th 2010 at 11:03 AM.
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  5. #5
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    x^2 -7x + 18x has to be greater than 0..
    as if its equated to 0, the roots are imaginary.
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  6. #6
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    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
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  7. #7
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by umangarora View Post

    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
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  8. #8
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    Quote Originally Posted by pencil09 View Post
    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 3or 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

    no, because you have not validated your two solutions from the quadratic in your initial inequality.

    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.
    Attached Thumbnails Attached Thumbnails inequalities-inequality.jpg  
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  9. #9
    Senior Member yeKciM's Avatar
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    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
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  10. #10
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    thanks for your answer....
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  11. #11
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    Quote Originally Posted by pencil09 View Post
    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.
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  12. #12
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    Quote Originally Posted by pencil09 View Post
    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
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