# Rearranging to quadratric equation

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• Oct 2nd 2012, 11:44 PM
yorkey
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? :-)
• Oct 3rd 2012, 12:13 AM
Prove It
Re: Rearranging to quadratric equation
Quote:

Originally Posted by yorkey
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 \displaystyle \begin{align*} x^4 \end{align*}. This will give

\displaystyle \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 \displaystyle \begin{align*} X \end{align*}, and use this to solve for \displaystyle \displaystyle \begin{align*} x\end{align*}.
• Oct 3rd 2012, 12:25 AM
yorkey
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?
• Oct 3rd 2012, 12:28 AM
Prove It
Re: Rearranging to quadratric equation
Quote:

Originally Posted by yorkey
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.