• Oct 2nd 2012, 11:44 PM
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? :-)
• Oct 3rd 2012, 12:13 AM
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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 \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*}.
• Oct 3rd 2012, 12:25 AM
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?
• Oct 3rd 2012, 12:28 AM
Prove It