4. Find the largest value of $3 k$ for which the following equation has a real root:
$$
\sqrt{x^{2}-k}+2 \sqrt{x^{3}-1}=x \text {. }
$$
4. Ans: 4
For the equation to have a solution, $k$ must be nonnegative, for if $kx \text {. }
$$

Rewrite the given equation as $2 \sqrt{x^{2}-1}=x-\sqrt{x^{2}-k}$. By squaring we obtain $2 x^{2}+k-4=-2 x \sqrt{x^{2}-k}$. Squaring again and solving for $x$, we obtain
$$
x^{2}=\frac{(k-4)^{2}}{8(2-k)} .
$$

Thus $0 \leq k<2$ and the only possible solution is $x=\frac{4-k}{\sqrt{8(2-k)}}$. Substitute this value of $x$ into the given equation and multiplying throughout by $\sqrt{8(2-k)}$, we obtain
$$
\sqrt{(k-4)^{2}-8 k(2-k)}+2 \sqrt{(k-4)^{2}-8(2-k)}=4-k .
$$

Hence, $|3 k-4|+2 k=4-k$ and so $|3 k-4|=-(3 k-4)$ which holds if and only if $3 k-4 \leq 0$; i.e. $3 k \leq 4$. Therefore, the maximum value of $3 k$ is 4 .