# Thread: I need help with another problem

1. ## I need help with another problem

The question came with the following instruction:
Express irrational solution in simplest radical form. If the equation has no solution, write "no solution".

(n-5)^2=20

Drakmord

2. Originally Posted by drakmord
The question came with the following instruction:
Express irrational solution in simplest radical form. If the equation has no solution, write "no solution".

(n-5)^2=20

Drakmord
Think of (n-5) as "x". If you had a problem like x^2=20, what would you do to solve it? You would take the square root of both sides. Same thing here. Make sense?

3. Originally Posted by Jameson
Think of (n-5) as "x". If you had a problem like x^2=20, what would you do to solve it? You would take the square root of both sides. Same thing here. Make sense?
So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

Am I close?
Thank you so much for your help Jameson!

Drakmord

4. Originally Posted by drakmord
So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

Am I close?
Thank you so much for your help Jameson!

Drakmord
Almost, but remember that when you take a square root there will be a positive root and a negative root. In other words $\displaystyle n-5 = \pm \sqrt{20}$

5. Originally Posted by drakmord
So (n-5)^2=20 would become: sqrt((n-5)^2)=Sqrt(20), which would then become: (n-5)=Sqrt(20)--> (n-5)=2(sqrt[5])?

Am I close?
Thank you so much for your help Jameson!

Drakmord
You are welcome. Remember to use the THANKS button when you find a helpful post.

6. Originally Posted by e^(i*pi)
Almost, but remember that when you take a square root there will be a positive root and a negative root. In other words $\displaystyle n-5 = \pm \sqrt{20}$
Many thanks for your help e^(i*pi)!
So it would come out to be $\displaystyle n-5 =2(\pm \sqrt{5})$?

drakmord

7. Originally Posted by drakmord
Many thanks for your help e^(i*pi)!
So it would come out to be $\displaystyle n-5 =2(\pm \sqrt{5})$?

drakmord
No because that says you should either add or subtract root 5 from 2 which is not the case. The plus or minus sign should be before the two because $\displaystyle \sqrt{20} = 2\sqrt{5}$ and $\displaystyle -\sqrt{20} = -(\sqrt{20}) = -(2\sqrt{5}) = -2\sqrt{5}$

8. Originally Posted by e^(i*pi)
No because that says you should either add or subtract root 5 from 2 which is not the case. The plus or minus sign should be before the two because $\displaystyle \sqrt{20} = 2\sqrt{5}$ and $\displaystyle -\sqrt{20} = -(\sqrt{20}) = -(2\sqrt{5}) = -2\sqrt{5}$

Ok, so the answer is:$\displaystyle (n-5)^2=20$ --> $\displaystyle \sqrt{(n-5)^2}=\sqrt{20}$--> $\displaystyle (n-5)=\sqrt{20}$ --> $\displaystyle (n-5)=\pm 2\sqrt{5}$ ?

Drakmord

9. Originally Posted by drakmord
Ok, so the answer is:$\displaystyle (n-5)^2=20$ --> $\displaystyle \sqrt{(n-5)^2}=\sqrt{20}$--> $\displaystyle (n-5)=\sqrt{20}$ --> $\displaystyle (n-5)=\pm 2\sqrt{5}$ ?

Drakmord
The $\displaystyle \pm$ should be in every step except the first one:

$\displaystyle (n-5)^2 = 20 \rightarrow \, n-5 = \pm \sqrt{20} \rightarrow \, n-5 = \pm 2\sqrt{5} \rightarrow \, n = 5 \pm 2\sqrt{5}$

I think I understand it now.

Drakmord

11. Originally Posted by drakmord