solving an equation algebraically

**alessandromangione** · February 21st 2010, 11:42 AM

x^4-3=2x+1
how can i solve it ? thanks in advance

**e^(i*pi)** · February 21st 2010, 11:55 AM

Originally Posted by alessandromangione

x^4-3=2x+1
how can i solve it ? thanks in advance

$x^4-2x-4=0$

I don't think you'll have much luck algebraically as both of the real roots appear to be transcendental.

**alessandromangione** · February 21st 2010, 12:01 PM

then i use the formula b^2 -square root of 4ac / 2a and i get x=0 and x=4...is it correct?

**e^(i*pi)** · February 21st 2010, 12:04 PM

Originally Posted by alessandromangione

then i use the formula b^2 -square root of 4ac / 2a and i get x=0 and x=4...is it correct?

That formula only works on quadratic equations and the equation in the original post is a quartic (x^4)

Did you mean $x^2-3 = 2x+1$ or $x^4-3 = 2x+1$

Given $ax^2+bx+c=0$ then the quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

**alessandromangione** · February 21st 2010, 12:50 PM

Originally Posted by e^(i*pi)

That formula only works on quadratic equations and the equation in the original post is a quartic (x^4)

Did you mean $x^2-3 = 2x+1$ or $x^4-3 = 2x+1$

Given $ax^2+bx+c=0$ then the quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

i mean x^4...so what formula can i use instead?

**e^(i*pi)** · February 21st 2010, 12:55 PM

Originally Posted by alessandromangione

i mean x^4...so what formula can i use instead?

You can't, at least not to my knowledge.

You'd have to use technology to get an iterative solution

Wolfram guesses too: http://www.wolframalpha.com/input/?i=x^4-3%3D2x%2B1+

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