Its part 1.....

Let $t=x^2$ . We have this equation:

$t^2+(1-2k)t+k^2-1=0$

The original equation has no roots when either of the following cases is correct

Case 1: The resultant equation has no roots:

$(1-2k)^2-4(k^2-1)<0$

$-4k+5<0$

$k>\displaystyle\frac{5}{4}$

Case 2: The resultant equation has negative roots $t_1,t_2$ (they needn't be distinct). This occurs if and only if their sum is negative and their product is positive, thus:

$\begin{array}{l} \left\{ \begin{array}{l} {(1 - 2k)^2} - 4({k^2} - 1) \ge 0 \\ 2k - 1 < 0 \\ {k^2} - 1 > 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} k \le \frac{5}{4} \\ k < \frac{1}{2} \\ k < - 1 \vee k > 1 \\ \end{array} \right. \\ k < - 1 \\ \end{array}$

Hence, the solution is $k \in \left( { - \infty , - 1} \right) \cup \left( {1,\displaystyle\frac{5}{4}} \right)$ , or AD.

The problem amounts to finding the values of $k$ for which $x^2=\frac{2k-1+\sqrt{-4k+5}}{2}$ and $x^2=\frac{2k-1-\sqrt{-4k+5}}{2}$ are either both imaginary or both less than zero. They are imaginary when the discriminant is negative. This leads to $k>\frac{5}{4}$ . They are both less than zero when we have both $2k-1+\sqrt{-4k+5}<0$ and $2k-1-\sqrt{-4k+5}<0$ . Solving both of these is a little bit trickier. First, rearrange the first one as $\sqrt{-4k+5}<1-2k$ . Then note that if $1-2k\leq0$ , the inequality is automatically false since the left side is always nonnegative. If $1-2k> 0$ , on the other hand, (or, equ., $k< \frac{1}{2}$ ), both sides are positive, so we can square each side of the inequality. So I do just that and proceed to solve: $-4k+5<1-4k+4k^2\Leftrightarrow k^2>1 \Leftrightarrow k<-1$ or $k>1$ . Combining those with $k<\frac{1}{2}$ results in $k<-1$ . Next, I analyze $2k-1-\sqrt{-4k+5}<0$ . Changing this to $2k-1<\sqrt{-4k+5}$ , I see that if $2k-1<0$ , the inequality is automatically $true$ . So part of the solution set of this inequality is $k<\frac{1}{2}$ . Now, assuming $2k-1\geq 0$ , I can square both sides and solve: $4k^2-4k+1<-4k+5\Rightarrow k^2<1\Rightarrow -1<k<1$ . Putting those together, I find the solution set of $2k-1<\sqrt{-4k+5}$ is $k<1$ . Therefore, both values of $x^2$ are negative iff $k<-1$ and $k<1$ (which simplifies to simply $k<-1$ ). So, in conclusion, there are no real roots iff both values of $x^2$ are imaginary or both values of $x^2$ are negative iff $k<-1$ or $k>\frac{5}{4}$ .