<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] Since $z\geq3$, we obtain:
$$\sqrt{3z+3}=\sqrt{2z+k}+\sqrt{3z-9}=\sqrt{5z+k-9+2\sqrt{(2z+k)(3z-9)}}=$$
$$=\sqrt{3z+3+k-6+2(z-3)+2\sqrt{(2z+k)(3z-9)}}\geq\sqrt{3z+3+k-6},$$
which gives that for $k>6$ our equation has no roots.
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