Answer to Question #149876 in Linear Algebra for Dhruv bartwal

Question

Find all the roots in C of the equation x^4-2x^3+4x^2+6x-21=0, give n that two of its roots are equal in magnitude and opposite in sign.

Answer 1

There is a general solution to the 4th power equation, but it is very complex. Fortunately this equation can be solved quite easily.

$x^4-2x^3+4x^2+6x-21=(x^2+a)(x^2+bx+c)=0$

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

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

$b=-2, ab=6,a=-3,c=4-a=4+3=7$

Therefore the equation is:

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

So either

$x=\pm\sqrt{3}$ or

$x^2-2x+7=0$

$D=b^2-4ac=4-28=-24,\sqrt{D}=2\sqrt{6}i$

$x=\frac{2\pm2\sqrt{6}i}{2}=1\pm\sqrt{6}i$

Out of 4 roots that we found $(\pm\sqrt{3},1\pm\sqrt{6}i)$ only the first two are equal in magnitude and opposite in sign, so n=1 (one pair).