Results 1 to 6 of 6

Math Help - Solving equations #1

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    27

    Red face Solving equations #1

    Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

    3) Factor completely. x - x^1/2 -6

    here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

    x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

    x^2+x-36 then from here i can factor it out... so im thinking this is the answer since
    (x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....

    4) ok last one( the one i cant seem to solve)

    \displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{  x-3}}=\displaystyle{\frac{x-2}{x-3}}

    so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

    i put everything on one side. then made it so each fraction had the same denominator so i got.

    -3+1-x-2/x-3 -> -4-x / x-3

    from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

    -(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

    help would be much appreciated! thanks!
    Last edited by mr fantastic; January 29th 2011 at 01:01 PM. Reason: Merged
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Red face

    Quote Originally Posted by Nismo View Post
    Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

    (3) Factor completely. x - x^1/2 -6

    here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

    x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

    x^2+x-36 then from here i can factor it out... so im thinking this is the answer since
    (x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....

    4) ok last one( the one i cant seem to solve)
    \displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{  x-3}}=\displaystyle{\frac{x-2}{x-3}}

    so i tried this problem about6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

    i put everything on one side. then made it so each fraction had the same denominator so i got.

    -3+1-x-2/x-3 -> -4-x / x-3

    from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

    -(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

    help would be much appreciated! thanks!
    For 3: Hint
    let y=x^{\frac{1}{2}} this gives
    y^2-y-6
    Last edited by mr fantastic; January 29th 2011 at 01:05 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    27
    Hey thanks for the help!

    for problem #3 im still a bit confused. Although i think i was able to solve the problem and got the answer of 2x-1 im not comprehending on how the steps where done. i do know and realize you substitued x^1/2 for y and solved it that way. but im puzzled as to what happened to that regular x did that get subsituted for the y^2 or somthing?
    Last edited by mr fantastic; January 29th 2011 at 01:04 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Nismo View Post
    4) ok last one( the one i cant seem to solve)

    \displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{  x-3}}=\displaystyle{\frac{x-2}{x-3}}

    so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

    i put everything on one side. then made it so each fraction had the same denominator so i got.

    -3+1-x-2/x-3 -> -4-x / x-3

    from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

    -(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

    help would be much appreciated! thanks!
    \displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{  x-3}}=\displaystyle{\frac{x-2}{x-3}}

    Multiply that through by the LCD [which is 2(x-3)] - this is to clear the denominator and give a linear equation

    \left(\dfrac{1}{2} \cdot 2(x-3)\right) + \left(\dfrac{1}{x-3} \cdot 2(x-3)\right) = \left(\dfrac{x-2}{x-3} \cdot 2(x-3)\right)


    You can simplify to give (x-3) + (2) = 2(x-2) which should be simple enough to solve

    edit: adding brackets
    Last edited by e^(i*pi); January 29th 2011 at 01:18 PM. Reason: I wonder how many words you could fit into this reason for editing box? It does seem rather a lot. Oh: see post for edit ;)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by Nismo View Post
    Hey thanks for the help!

    for problem #3 im still a bit confused. Although i think i was able to solve the problem and got the answer of 2x-1 im not comprehending on how the steps where done. i do know and realize you substitued x^1/2 for y and solved it that way. but im puzzled as to what happened to that regular x did that get subsituted for the y^2 or somthing?
    (2x-1) is not the answer...

    As TheEmptySet has suggested:

    y^2-y-6 can be factorised as:

    y^2-3y+2y-6\;=\;......

    for your question no. 4. start by multiplying both sides of the equation by x-3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,952
    Thanks
    1629
    Quote Originally Posted by Nismo View Post
    Hi i need some help please on how to solve these problems. i tried each one but either cant figure out the answer/ i got an answer but im not sure if its correct.

    3) Factor completely. x - x^1/2 -6

    here i changed the -x^1/2 into square root x since the two mean the same thing. so then i had.

    x-(squareroot x) -6. i then to kill the square root i rooted the whole problem so then it came out to be.

    x^2+x-36 then from here i can factor it out
    "Rooted the whole problem"? If you mean by that x-\sqrt{x}- 6= \sqrt{x^2+ x- 36},that is simply not true. The square root of a^2- b^2 NOT equal to a- b. For example, 5^2- 4^2= 25- 16= 9 and \sqrt{9}= 3, not 5- 4= 1.

    ... so im thinking this is the answer since
    (x+6)(x-6) wotn work or even changing it out with 9 and 4 or 18 and 2... if you know what i mean...i could complete the square but then again im factoring it out not solving for x so like i said im not sure if that is the answer....
    Instead, let y= \sqrt{x} so that x= y^2. Now you have y^2- y- 6= (y- 3)(y+ 2)= 0. Setting y= \sqrt{x}= x^{1/2} again, that tells us that x- x^{1/2}- 6= (x^{1/2}- 3)(x^{1/2}+ 2).

    Once again, \sqrt{a+ b} is NOT \sqrt{a}+ \sqrt{b}. That was your mistake.


    4) ok last one( the one i cant seem to solve)

    \displaystyle{\frac{1}{2}}+\displaystyle{\frac{1}{  x-3}}=\displaystyle{\frac{x-2}{x-3}}

    so i tried this problem about 6 diffrent ways and ended up with ansers like 3,2,4,9,7 and when i plug them in they dont equal out..... i tell you one way i tried this..

    i put everything on one side. then made it so each fraction had the same denominator so i got.

    -3+1-x-2/x-3 -> -4-x / x-3

    from here i think i loose it and i multiply the top and bottom by (x-3) and end up getting

    -(x^2+2x-12) / x^2-6x+9 i try to factor this out but only the bottom i can factor out and not the top...

    help would be much appreciated! thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 01:41 AM
  2. Solving some equations
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 24th 2010, 08:52 AM
  3. Solving these equations...
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 27th 2009, 05:06 AM
  4. solving equations help
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 17th 2009, 05:09 AM
  5. Solving Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 24th 2008, 10:38 AM

Search Tags


/mathhelpforum @mathhelpforum