# Thread: solving an equation algebraically

1. ## solving an equation algebraically

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

2. 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.

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

4. 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 $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

5. 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?

6. 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+