Question

If the sum of two roots of $x^4-2x^3+4x^2+6x-21=0$ is zero, then which of the following is/are true?

• A. one of the roots of the equation is $1+i\sqrt{6}$
• B. all roots of the equation are real
• C. the equation has only two real roots
• D. sum of all the real roots of the equation is $0$

Answer 1

Answer: (A), (C), (D)

sum of all the real roots of the equation is $0$

$x^4-2x^3+4x^2+6x-21=0$
Let $\alpha ,\beta ,\gamma$ and $\delta$ be the roots.
Then $S_1=\alpha +\beta +\gamma +\delta =2$
$S_2=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =4$
$S_3=\alpha \beta \gamma +\alpha \beta \delta +\beta \gamma \delta +\alpha \gamma \delta =-6$
$\alpha \beta \gamma \delta =-21$

Let $\alpha +\beta =0\cdots \left(1\right)$
$\Rightarrow \gamma +\delta =2\cdots \left(2\right)$

$\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =4$
$\Rightarrow \alpha \beta +\alpha \gamma +\alpha \delta -\alpha \gamma -\alpha \delta +\gamma \delta =4[\because \beta =-\alpha ]$
$\Rightarrow \alpha \beta +\gamma \delta =4\cdots \left(3\right)$

$\alpha \beta \gamma +\alpha \beta \delta +\beta \gamma \delta +\alpha \gamma \delta =-6$
$\Rightarrow \alpha \beta \gamma +\alpha \beta \delta -\alpha \gamma \delta +\alpha \gamma \delta =-6[\because \beta =-\alpha ]$
$\Rightarrow \alpha \beta (\gamma +\delta )=-6$
$\Rightarrow \alpha \beta =-3\cdots \left(4\right)$

$\alpha \beta \gamma \delta =-21$
From $\left(4\right),\gamma \delta =7\cdots \left(5\right)$

Solving $\left(1\right)$ and $\left(4\right)$, we get
$\alpha =\sqrt{3},\beta =-\sqrt{3}$ or $\alpha =-\sqrt{3},\beta =\sqrt{3}$

Solving $\left(2\right)$ and $\left(5\right)$, we get
$\gamma =1+i\sqrt{6},\delta =1-i\sqrt{6}$ or $\gamma =1-i\sqrt{6},\delta =1+i\sqrt{6}$

$\therefore$ Roots are $\sqrt{3},-\sqrt{3},1+i\sqrt{6},1-i\sqrt{6}$