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

$$(1)$$ 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

• For the first factor, the discriminant is $D_1=k^2-4(3)(3)=k^2-36$. For the second factor, $D_2=k^2-4(1)(-1)=k^2+4$. – Ángel Mario Gallegos Nov 1 '14 at 4:40

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 $D_1\le0$