real value of $k$ inirrational equation

Question

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

Answer 1

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.

Answer 2

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 $k$.

Answer 3

Hint : From the start : $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ leads to $$ 3z+3 +3z-9-2\sqrt{3z+3}\sqrt{3z-9}=2z+k $$ so $$2\sqrt{3z+3}\sqrt{3z-9}=4z-6-k $$ and squaring again to $$ 4(3z+3)(3z-9)=(4z-6-k)^2 $$ so $$20z^2 +4(48+2k)z-27\times 4 -(6+k)^2=0 \tag{3} $$ You won't have solutions if $$ \Delta_k= (4(48+2k))^2-4\times20\times -(27\times 4 +(6+k)^2 < 0 $$ With gives you a domain for $k$, by studying this last second degree equation in $k$.

But, since we took the power 2 twice, a solution of (3) may not be a solution of (1) : so it remains to verify if $k$ is such (3) has a solution ($\Delta_k\geq 0$), the quantities before squaring remain positive ; if not you can add those values of $k$ in the set of case where (1) has no solution.