Question

Solve the equation $x^4-4x^2+8x+35=0$ having given that one root is $2+\sqrt{-3}.$

• A. The roots of the given equation are $-2\pm i,2\pm \sqrt{3}i.$
• B. The roots of the given equation are $2\pm i,2\pm \sqrt{3}i.$
• C. The roots of the given equation are $\pm 2+i,2\pm \sqrt{3}i.$
• D. The roots of the given equation are $\pm 2-i,2\pm \sqrt{3}i.$

Answer 1

The correct option is A The roots of the given equation are $-2\pm i,2\pm \sqrt{3}i.$

The equation is $x^4-4x^2+8x+35=0.$ ...(1)
One root of this equation is given as $2+\sqrt{3}i.$
Since the complex roots occur in conjugate pairs, the other root must be $2-\sqrt{3}i.$
$\therefore S=4,p=7$The quadratic factor corresponding to these two roots is $x^2-Sx+P$ or $x^2-4x+7.$
Then the other quadratic factor of L.H.S. of (1)
is of the form $x^2+px+5.$
Hence we have the identity
$x^4-4x^2+8x+35=(x^2-4x+7)(x^2+px+5).$
Equating the coefficient of x on both sides of the above identity, we get $8=7p-20$ or $p=4$.
[Note that same value of, will be obtained by equating the coefficient of $x^2$].
Hence the other two roots of the equation are the roots of the equation $x^2+4x+5=0$
or $x=\frac{-4\pm \sqrt{16-20}}{2}=\frac{-4\pm \sqrt{-4}}{2}=-2\pm i.$
Ans: A