distinct possible real roots of the equation >$(3x^2+kx+3)(x^2+kx-1) = 0.$

Question

$(1)$ If $k\in\mathbb{R}$. Then minimum no. of distinct possible real roots of the equation

$(3x^2+kx+3)(x^2+kx-1) = 0.$

$\bf{My\; Try::}$ Discriminant of quadratic equation $3x^2+kx+3 = 0$ is $D_{1} = (k^2-36)$

$\bullet $ If $|k|<6\:,$ Then $D_{1}<0\;\;,\bullet\; $ If $|k|=6,$ Then $D_{1}=0\;,$ If $|k|>6\;,$ Then $D_{1}>0$

And Discriminant of quadratic equation $x^2+kx-1 = 0$ is $D_{2} = (k^2+4)>0\;\forall \;k\;\in \mathbb{R}$

So we can conclude that $(3x^2+kx+3)(x^2+kx-1)=0$ has Max.$2$ distinct real roots, If $|k|>6$

and $2$ equal roots, If $|k|=6.$

and no real roots , If $|k|<6.$

So Min. possible distinct real roots of the equation $(3x^2+kx+3)(x^2+kx-1)=0$ is $=0$

But answer given as $ = 2$

plz help me, Thanks

Answer 1

As $D_2=k^2+4\ge4>0$ for real $k$ $$x^2+kx-1=0$$ will always have distinct real roots

For $D_1=k^2-4\cdot3\cdot3=k^2-36,$

$$3x^2+kx+3=0$$ won't have distinct real roots if $D_1\le0$