Results 1 to 5 of 5

Math Help - Roots of equation

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    21

    Question Roots of equation

    I've gone through my workings twice and re-did, but somehow I keep getting an irregularity.

    Quote Originally Posted by Question
    The roots of the equation x^2 - (q - 5)x = \frac{q}{2} are \alpha and \alpha+5. Find the possible values of q.
    I did it like this.

    Since the roots are \alpha and \alpha + 5, I substituted them into the equation.

    \alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)
    \alpha^2 - (q - 5)\alpha = \alpha^2 + 10\alpha + 25 - (q\alpha - 5\alpha + 5q - 25)
    \alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha + b\alpha  - 5q + 25)
    [tex]\alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)[tex]

    Therefore -(q - 5)\alpha = (15 - q)\alpha, 50 - 5q = 0
    5 - q = 15 - q, q = 10

    But 5-q = 15-q doesn't even make any sense.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by fterh View Post
    I've gone through my workings twice and re-did, but somehow I keep getting an irregularity.



    I did it like this.

    Since the roots are \alpha and \alpha + 5, I substituted them into the equation.

    \alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)
    \alpha^2 - (q - 5)\alpha = \alpha^2 + 10\alpha + 25 - (q\alpha - 5\alpha + 5q - 25)
    \alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha + b\alpha  - 5q + 25)
    [tex]\alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)[tex]

    Therefore -(q - 5)\alpha = (15 - q)\alpha, 50 - 5q = 0
    5 - q = 15 - q, q = 10

    But 5-q = 15-q doesn't even make any sense.
    There are some typos in your work so I don't know what you meant to write. But I think starting with

    \alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)

    while valid results in a loss of information, because the c term in ax^2+bx+c=0 gets canceled out. When I worked out everything according to your first step, I got q=2\alpha+10, which is a valid restriction on q but is not the solution set. (Because not all values of \alpha are possible.)

    To solve, I did:

    \displaystyle x^2-(q-5)x-\frac{q}{2}=0

    \displaystyle x=\frac{q-5\pm\sqrt{(q-5)^2-4(-\frac{q}{2})}}{2}

    \dots

    \displaystyle \frac{q-5-\sqrt{q^2-8q+25}}{2}+5 = \displaystyle \frac{q-5+\sqrt{q^2-8q+25}}{2}

    \dots

    10=2\sqrt{q^2-8q+25}

    \dots

    q\in\{0,8\}

    And also, the question doesn't ask it, but \alpha\in\{-5,-1\}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2010
    Posts
    21
    Sorry but can you highlight to me where are the typos? I looked through it and it seems to be in order except for a trailing / at the 3rd last statement.

    So you mean that my method won't work?
    Last edited by fterh; June 19th 2010 at 10:13 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by fterh View Post
    Sorry but can you highlight to me where are the typos? I looked through it and it seems to be in order except for a trailing / at the 3rd last statement.
    Unfortunately the \color{} command doesn't work like it did in the old forum, so I can't highlight directly, but in the line you wrote

    \alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha +  b\alpha  - 5q + 25)

    there is a misplaced "|" symbol and also the letter b out of the blue, and a stranded end parenthesis at the very end, then the next line isn't rendered in LaTeX due to the missing "/" .... well I can work out what you mean with some effort, but really, why should I put in all the effort to read what you didn't put much effort into writing/editing?

    So I can see you go from

    \alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)

    to

    - (q - 5) = 15 - q\quad\text{AND}\quad 0 = 50 - 5q

    This doesn't work because consider a different example:

    10x = 5x + 12

    Which by your step would lead to

    10 = 5\quad\text{AND} \quad0 = 12

    So you got confused with this kind of situation:

    If a_1x^2 + b_1x + c_1 and a_2x^2 + b_2x + c_2 are the same polynomial, then a_1 = a_2, b_1 = b_2, and c_1 = c_2.

    Quote Originally Posted by fterh View Post
    So you mean that my method won't work?
    Well, like I said, using your method results in q=2\alpha+10. You would then have to go back and substitute

    \displaystyle x-(q-5)x=\frac{q}{2}

    \displaystyle \alpha-((2\alpha+10)-5)\alpha=\frac{2\alpha+10}{2}

    and solve for \alpha, etc. Seems like more work, overall.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2010
    Posts
    21
    I see, thanks for your help

    Really sorry about the typos, I must have been blind
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Can this equation have any other roots
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: October 10th 2011, 05:26 AM
  2. Roots on an equation
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: September 30th 2010, 05:45 AM
  3. Roots of an equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 22nd 2009, 07:31 AM
  4. Four roots of an equation
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 9th 2008, 10:44 AM
  5. roots of an equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 9th 2007, 08:22 AM

Search Tags


/mathhelpforum @mathhelpforum