Question

The product of uncommon real roots of the equation $x^4+2x^3-8x^2-6x+15=0$ and $x^3+4x^2-x-10=0$ is

• A. $6$
• B. $-6$
• C. $8$
• D. $-8$

Answer 1

The correct option is A $6$

Let $\alpha$ be the common root.
$\therefore {\alpha }^4+2{\alpha }^3-8{\alpha }^2-6\alpha +15=0$
${\alpha }^3+4{\alpha }^2-\alpha -10=0$
$\therefore$ Given equations are $({\alpha }^2-3)({\alpha }^2+2\alpha -5)=0$ &
$(\alpha +2)({\alpha }^2+2\alpha -5)=0$
$\Rightarrow$ For common roots ${\alpha }^2+2\alpha -5=0$
$\therefore$ Product of uncommon real roots is $(-3)(-2)=6$.