Answer:
No solution
Step-by-step explanation:
Given that,
[tex]3\sqrt{x+2}+\sqrt{x-4}=0[/tex]
Box x under square root. Inside the square root number must be positive.
Therefore,
[tex]x+2\geq 0[/tex]
[tex]x\geq -2[/tex]
[tex]x-4\geq0[/tex]
[tex]x\geq4[/tex]
[tex]3\sqrt{x+2}+\sqrt{x-4}=0[/tex]
[tex]3\sqrt{x+2}=-\sqrt{x-4}[/tex]
Taking square both sides
[tex](3\sqrt{x+2})^2=(-\sqrt{x-4})^2[/tex]
[tex]9(x+2)=(x-4)[/tex]
[tex]9x+18=x-4[/tex]
[tex]9x-x=-4-18[/tex]
[tex]8x=-22[/tex]
[tex]x=-\dfrac{11}{4}[/tex]
Now take intersection of all three solution.
[tex]x=-\dfrac{11}{4}[/tex],[tex]x\geq4[/tex],[tex]x\geq -2[/tex]
No intersection value of x
No solution
