Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!

Thanks in advance

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- June 2nd 2011, 04:56 AMtheleontanSolving a Quartic
Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!

Thanks in advance - June 2nd 2011, 05:06 AMskeeter
- June 2nd 2011, 05:23 AMProve It
If you were not able to recognise what Skeeter posted straight away, notice that if you let , then your equation becomes

which is an easy quadratic to solve. Once you have you can find . - June 2nd 2011, 05:05 PMtheleontan
- June 2nd 2011, 05:37 PMWilmer
- June 2nd 2011, 05:37 PMscounged
Let's look at your original equation:

note that

This means that your original equation can be rewritten as

What Prove It did was to substitute for .

If you substitute for in the equation above,

you will get this:

This equation can be solved for with the quadratic formula, as it's a quadratic equation with as an unknown, and as we know that stands for , this means that If we know we can get that , and then solve for when we have figured out . - June 3rd 2011, 04:35 AMjgv115
This is a longer method but the more ways you know how to do something the better! Alternatively, the factor theorem could be used.

Subbing x = 1 into the quartic gives you 0 so x-1 is a factor. Using long division we can get a cubic answer. Using the factor theorem again on that will get you a factorisable quadratic.

Then you will have 4 factors!