Solve #x^4-2x^3+4x^2+6x-21=0#?

Question

4260 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-25T11:45:22

The roots are #x=+-sqrt(3), 1+-sqrt(6)i#

Quartic equations are quite difficult to solve. There is a general solution to the quartic, but it is very complex.

Fortunately this equation can be solved quite easily.

We want to solve #x^4-2x^3+4x^2+6x-21=0#

Assume that we can factorise the equation into the product of two quadratics.

#(x^2+a)(x^2+bx+c) = 0#

Multiply the two quadratics.

#x^4+bx^3+(a+c)x^2+abx+ac=0#

Comparing terms gives.

#b=-2, a+c=4, ab=6, ac=-21#

We have #b=-2#, #ab=6# means that #a=-3#.
Now #a+c=4# means #c=7#.
Verify with the last term #ac=-3*7=-21#.

So the equation factorises to.

#(x^2-3)(x^2-2x+7)=0#

Factorise each term to give the roots.

#x^2-3=0#, is #x^2=3#, roots #x=+-sqrt(3)#

#x^2-2x+7=0# complete the square #(x-1)^2=-6# gives complex roots #x=1+-sqrt(6)i#