doubts in equations

$\underline{\underline{Ans:(1)}}$\Rightarrow$ Let $x$ be a No. Then Its Reciprocal is $\frac{1}{x}$ and $x\neq0$\\\\ Then let $k=x+\frac{1}{x}$ \\\\ $\underline{\underline{Case:(1):}}\Rightarrow$ If $x>0$ Then \\\\ $k=x+\frac{1}{x}\geq 2\sqrt{x.\frac{1}{x}}\geq 2.$(Using A.M\geq G.M).$\\\\ $So $\boxed{k\geq2}$\\\\ $\underline{\underline{Case:(2):}}\Rightarrow$ If $x<0$ Then \\\\ $k = x+\frac{1}{x}<=-2\sqrt{x.\frac{1}{x}}\leq -2.$(Using A.M\geq G.M).$\\\\ $So $\boxed{k\leq-2}$\\\\ So from These Condition $\textcolor[rgb]{0.,1.,0.}{\boxed{\boxed{\textcolor[rgb]{1.,0.,0.}{k\in(-\infty,-2]\cup[2,+\infty)}}}}$

2) For question 2's 2nd part, they cannot have a common non-real root because if there is a common non real root then both the roots must be common as roots occur in conjugate pairs.

Now in that case,

k(3x²-2mx-4)=x²-4mx+2

Comparing coefficients,

3k=1 or k=1/3

-2mk=-4m or k=2

Hence, this is not possible.

For the 1st part, get the discriminants of both equations and check for two cases,

i) Roots are rational

ii) Roots are irrational (then both roots will be common).

I think you should get the required condition.