Results 1 to 7 of 7

Math Help - Equation

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    56

    Equation

    Equation-dsc00539.jpg
    For question 21(1), i tried many times but i couldn't show it, after substituting everything, the equation appears to be really complicated, i wonder can anyone help me with 21(i)? Thanks in advance...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by MichaelLight View Post
    Click image for larger version. 

Name:	DSC00539.jpg 
Views:	43 
Size:	433.4 KB 
ID:	20979
    For question 21(1), i tried many times but i couldn't show it, after substituting everything, the equation appears to be really complicated, i wonder can anyone help me with 21(i)? Thanks in advance...

    Use Viette's formulae: " the product of both roots of a quadratic eq. equals its free coefficient divided by its leading coef.,

    and the sum of the roots equals minus its linear coef. divided by its leading one".

    Thus, \displaystyle{\frac{m^2}{ln}=\frac{\left(l\lambda(  \lambda+1)^2+(\lambda+1)^2\right)^2}{ln}=\frac{l(\  lambda+1)^6}{n}}

    \displaystyle{\frac{b^2}{ac}=\frac{\left(-a(\alpha+\beta)\right)^2}{a^2\alpha\beta}=\frac{(\  alpha+\beta)^2}{\alpha\beta}=\frac{(\lambda\beta+\  beta)^2}{\lambda\beta^2}=\frac{(\lambda+1)^2}{\lam  bda}}.

    Well, now just show that both expressions in the RHS above are equal (Viete again),

    and for part (ii) is similar.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    \displaystyle ax^2 + bx + c = 0

    Thus
    \displaystyle a \left ( x^2 + \frac{b}{a}x + \frac{c}{a} \right ) = 0

    If the two roots are \beta and \lambda \beta then we know that
    \displaystyle a ( x - \beta ) (x - \lambda \beta ) = 0

    Expanding this and equating powers of x from your a, b, c equation form gives

    \displaystyle \frac{b}{a} = - \beta ( \lambda + 1 )

    and
    \displaystyle \frac{c}{a} = \lambda \beta ^2

    Solve the first of these for b and square it:
    \displaystyle b^2 = a^2 \beta ( \lambda + 1 )^2

    So
    \displaystyle \frac{b^2}{a} = a \beta ( \lambda + 1 )^2

    and since
    \displaystyle c = a \lambda \beta ^2

    we have, after some simplifying:
    \displaystyle \frac{b^2}{ac} = \frac{( \lambda + 1 )^2}{\lambda}

    You do the l, m, n quadratic the same way.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2011
    Posts
    56
    Okay... i solved part (i) but i face difficulty in solving (ii) now... please help...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    According to the properties of quadratic equation, α + β = -p and αβ = q.

    Then (α + β)^2 = .......

    (α - β)^2 = (α + β)^2 - 4αβ = ........

    (α^2 - β^2) = .........
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2011
    Posts
    56
    Quote Originally Posted by sa-ri-ga-ma View Post
    According to the properties of quadratic equation, α + β = -p and αβ = q.

    Then (α + β)^2 = .......

    (α - β)^2 = (α + β)^2 - 4αβ = ........

    (α^2 - β^2) = .........
    I don't understand... why α + β = -p and αβ = q..? Isn't the given roots are (α^2 + β^2) and (α^2 - β^2)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    If α and β are the roots of ax^2 + bx + c = 0, then the sum of the roots = α + β = - b/a and product of the roots = αβ = c/a.

    In the second equation,
    (α^2 + β^2) + (α^2 - β^2) = -p and (α^2 + β^2) * (α^2 - β^2) = q.....(1)

    Now (α^2 + β^2) = (α + β)^2 - 2αβ = ...?

    (α - β)^2 = (α + β)^2 - 4αβ =.......? Find (α - β)

    (α^2 - β^2) = (α + β)(α - β) = ......

    Substitute these values in eq.(1)
    Last edited by sa-ri-ga-ma; February 27th 2011 at 10:19 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 11th 2011, 02:17 AM
  2. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 12:39 PM
  3. Replies: 2
    Last Post: May 18th 2009, 01:51 PM
  4. Replies: 2
    Last Post: April 28th 2009, 07:42 AM
  5. Replies: 1
    Last Post: October 23rd 2008, 04:39 AM

Search Tags


/mathhelpforum @mathhelpforum