Results 1 to 4 of 4

Math Help - Rearranging to quadratric equation

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    93

    Rearranging to quadratric equation

    Here I've got a sum that I have to reorder into a quadratic equation so I can find the two roots of it, but I'm struggling to understand how. Once into the quadratic form, I can easily find the roots, but the problem for me is getting it into the quadratic form.

    18/x^4 + 1/x^2 = 4

    Now what I did was put the x^4 on the other side. So it's:

    x^4 - x^-2 - 18 = 0

    Is this right?

    According to the mark scheme, I had to add the two fractions together, but I don't see why that is the right option. Any help? :-)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Rearranging to quadratric equation

    Quote Originally Posted by yorkey View Post
    Here I've got a sum that I have to reorder into a quadratic equation so I can find the two roots of it, but I'm struggling to understand how. Once into the quadratic form, I can easily find the roots, but the problem for me is getting it into the quadratic form.

    18/x^4 + 1/x^2 = 4

    Now what I did was put the x^4 on the other side. So it's:

    x^4 - x^-2 - 18 = 0

    Is this right?

    According to the mark scheme, I had to add the two fractions together, but I don't see why that is the right option. Any help? :-)
    Not quite. What you need to do is get rid of the fractions. To do this, multiply both sides by the largest order denominator, in this case, \displaystyle \begin{align*} x^4 \end{align*}. This will give

    \displaystyle \begin{align*} \frac{18}{x^4} + \frac{1}{x^2} &= 4 \\ x^4 \left( \frac{18}{x^4} + \frac{1}{x^2} \right) &= 4x^4 \\ 18 + x^2 &= 4x^4 \\ 0 &= 4x^4 - x^2 - 18 \\ 0 &= 4X^2 - X - 18 \textrm{ if we let } X = x^2 \end{align*}

    Now solve for \displaystyle \begin{align*} X \end{align*}, and use this to solve for \displaystyle \begin{align*}  x\end{align*}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    93

    Re: Rearranging to quadratric equation

    It's strange, I don't know this rule. Sorry to keep you, but could you just lay out the whole "multiply both sides by the lowest order denominator" in a more general way so I can remember it?

    So: if I have 2 denominators with variables of different exponents, I multiply the entire LHS and the entire RHS by this value, and then simplify. Is that it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Rearranging to quadratric equation

    Quote Originally Posted by yorkey View Post
    It's strange, I don't know this rule. Sorry to keep you, but could you just lay out the whole "multiply both sides by the lowest order denominator" in a more general way so I can remember it?

    So: if I have 2 denominators with variables of different exponents, I multiply the entire LHS and the entire RHS by this value, and then simplify. Is that it?
    Well the reason is because if you were to try to add the fractions, you need a common denominator. Then once they're added, to simplify so that you can solve the equation, you need to multiply both sides by the denominator. Try it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. rearranging an equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 14th 2013, 09:42 AM
  2. Need help rearranging an equation
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 22nd 2010, 09:41 AM
  3. Rearranging Equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 15th 2008, 07:20 PM
  4. Rearranging an equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 16th 2008, 12:44 PM
  5. Equation rearranging?
    Posted in the Pre-Calculus Forum
    Replies: 14
    Last Post: March 11th 2008, 07:57 AM

Search Tags


/mathhelpforum @mathhelpforum