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

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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 $x=1+-sqrt(6)i$

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