Question

$x^4-2x^3+4x^2+6x-21=0$ if two of its roots are equal in magnitude but opposite in sign.Then among the 4 roots the real roots be $\pm k$.Then find $k^2$ ?

Answer 1

$\alpha +\beta =0but\alpha +\beta +\gamma +\delta =2$
$\therefore \gamma +\delta =2.$
Let $\alpha \beta =pand\gamma \delta =q$
$\therefore$ Given equation is equivalent to
$(x^2+p)(x^2-2x+q)=0.$
Comparing the coefficients, we get
$p+q=4,-2p=6,pq=-21$
$\therefore p=-3,q=7$
Thus equation reduced to,
$(x^2-3)(x^2-2x+7)=0$
Hence roots are $\pm \sqrt{3},$ and $1\pm i\sqrt{6}.$