Consider the two quartic equations below.

Which of the two equations is easier to solve? And why?

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- January 27th 2008, 04:44 AMJaneFairfaxWhich equation is easier to solve?
Consider the two quartic equations below.

Which of the two equations is easier to solve? And why? - January 27th 2008, 04:57 AMSean12345
At first glance the second one as you can take a factor of x out straight away however the cubic that is left; is quite difficult to find a factor of.

The first one however is much easier even though at first look it doesn't seem to be. This is because a factor of can be noticed relatively quickly leaving the cubic in which the factor can be found leaving the quadratic . - January 27th 2008, 05:01 AMCaptainBlack
Well by inspection is a factor of the first, as is so:

So the first is no more difficult to solve than a quadratic.

The second has as a factor leaving a cubic, which by applying the rational

root theorem we discover has a factor , so:

So the second is also no more difficult to solve than a quadratic.

So they are equally easy to solve.

(I should add that the two quadratic factors do have real roots but they are not rational)

RonL - January 27th 2008, 07:41 AMJaneFairfax
Thanks! :D

This was the result of trying to solve a problem posed by someone on another site. We were to solve the following equation:

The idea of course is NOT to tackle the equation directly by expanding everything out into a mess, but to make a substitution to simplify things a bit. I suggested the substitution , which would cause the constant terms to disappear, leading to (which is Eqn (2) above).

But someone else suggested the substitution instead. This leads to Eqn (1): . At first glance, I was like Sean, thinking that my equation with constant term zero was easier to tackle. Now I know, thanks to CaptainBlack, that both subsitutions are equally good. (Handshake)

What I am not happy with is that the other person said my substitution was “stupid” when his was at best only as good as mine: http://www.mathisfunforum.com/viewtopic.php?id=9144 :mad: - January 27th 2008, 08:56 AMCaptainBlack
Immature f**l, because he has a solution he thinks it is the best.

Possibly he does not have the tools to easily spot rational roots other than +/-1.

Of course he may be refering to it being less work to expand a cube of the form

than to expand a fourth power like

RonL - January 27th 2008, 12:24 PMSean12345
It amazes me as to how ignorant some people are; they think you are stupid because you use a different method to them. As long as the method is correct and leads to the correct solution how can they make such a remark?

- January 27th 2008, 02:24 PMmr fantastic
hmmmmmph .... mathisfun indeed ..... until someone loses an eye (no offence to Euler).

Some salient points:

1. The other person is a*novice*. You're a*super duper member*- with all the rep and cred that brings with it.

2. By their own signature, their English is not very good. Maybe they meant the word*smart*but got all muddled up. Maybe they thought it funny --> fun, like the forum name - in which case, they need to add 'my sense of humour is not very good, either' to their signature.

3. The other person posed the original question. Clearly*your*solution suggested a similar approach to them - unfortunately it also gave them delusions of grandeur.

4. The good Captain, as always, raises some good points:

5. At the end of the day, the mathematics does the talking. It says that neither solution is easier than the other. And that's that.

6. Get some perspective - Read the words to The Sunscreen Song :cool: (they're at the end).