<s>[INST] real value of $k$ inirrational equation Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.
solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$
$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$
$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$
$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$
$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$
Help me how   i solve it after that point 
 [/INST] Starting from your last step,
$$4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$$
is equivalent to
$$28z^2+(24+12k+48+4k)z-(k^2+12k-144)=0  $$
and this as no solution (since $z$ is real, for the squared roots be defined) if and only if
$$ \Delta = (24+12k+48+4k)^2-4\times 28 \times -(k^2+12k-144) <0 $$
So it remain to study a 2de order equation degree in $k$.
</s>