1 Answer

Azizjon Kholmatov · Answer 1 · 2023-08-21T13:10:07+0000

To solve the expression:

$2\cdot\sqrt[]{n}=n-3$

We need to square to both sides of the equation. Then, we have:

$(2\cdot\sqrt[]{n})^2=(n-3)^2\Rightarrow4\cdot n=(n-3)^2\Rightarrow4n=n^2-6n+9_{}$

Then,

$n^2-6n+9-4n=0\Rightarrow n^2-10n+9=0$

Then, we have that the solutions for this quadratic equation are: n = 1, and n = 9, since

$(n-1)\cdot(n-9)=0,n=1,n=9$

And

$(n-1)\cdot(n-9)=n^2-10n+9$

We need to check the results. For n = 1:

$2\sqrt[]{1}=1-3\Rightarrow2\cdot1=-2\Rightarrow2\\e-2$

Then, for n = 1, it is not a solution.

We need to check for n = 9:

$2\cdot\sqrt[]{9}=9-3\Rightarrow2\cdot3=6\Rightarrow6=6$

Then, the solution for the expression above 2*sqrt(n) = n - 3 is n = 9.

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