To solve the expression:
[tex]2\cdot\sqrt[]{n}=n-3[/tex]We need to square to both sides of the equation. Then, we have:
[tex](2\cdot\sqrt[]{n})^2=(n-3)^2\Rightarrow4\cdot n=(n-3)^2\Rightarrow4n=n^2-6n+9_{}[/tex]Then,
[tex]n^2-6n+9-4n=0\Rightarrow n^2-10n+9=0[/tex]Then, we have that the solutions for this quadratic equation are: n = 1, and n = 9, since
[tex](n-1)\cdot(n-9)=0,n=1,n=9[/tex]And
[tex](n-1)\cdot(n-9)=n^2-10n+9[/tex]We need to check the results. For n = 1:
[tex]2\sqrt[]{1}=1-3\Rightarrow2\cdot1=-2\Rightarrow2\ne-2[/tex]Then, for n = 1, it is not a solution.
We need to check for n = 9:
[tex]2\cdot\sqrt[]{9}=9-3\Rightarrow2\cdot3=6\Rightarrow6=6[/tex]Then, the solution for the expression above 2*sqrt(n) = n - 3 is n = 9.
