# Thread: What is the clue to think that way?

1. ## What is the clue to think that way?

In the "Foundation Maths" book in chapter (19): "The exponential function", there was an example solving the following problem (I used "^" to denote power):

(e^2t + 2e^2 + 1)^1/2

In solving the problem, this was converted to ((e^t + 1)^2)^1/2

What clue can tell that I have to solve the problem this way and not by just taking the root of every element?

Thanks.

2. Originally Posted by SWEngineer
In the "Foundation Maths" book in chapter (19): "The exponential function", there was an example solving the following problem (I used "^" to denote power):

(e^2t + 2e^2 + 1)^1/2

In solving the problem, this was converted to ((e^t + 1)^2)^1/2

What clue can tell that I have to solve the problem this way and not by just taking the root of every element?

Thanks.
$\sqrt{e^{2t}+2e^2+1}=\sqrt{\left(e^t+1\right)\left (e^t+1\right)}$

Taking the square root of each term does not work.

$\sqrt{4}+\sqrt{4}=2+2=4$

$\sqrt{4+4}=\sqrt{8}=\sqrt{(4)2}=\sqrt{4}\sqrt{2}=2 \sqrt{2}<4$

Or

$\sqrt{8}<\sqrt{9}\Rightarrow\ \sqrt{8}<3<4$

To take a square root, you can do so when you are taking the square root of a square.

$2^2=4\Rightarrow\sqrt{4}=2$

$3^2=9\Rightarrow\sqrt{9}=3$

$3^2+4^2=5^2\Rightarrow\sqrt{3^2+4^2}=\sqrt{5^2}=5$

$\left(e^t+1\right)\left(e^t+1\right)=e^t\left(e^t+ 1\right)+1\left(e^t+1\right)=e^{2t}+e^t+e^t+1=e^{2 t}+2e^t+1$

However, this is a square

$e^{2t}+2e^t+1=\left(e^t+1\right)^2$

3. Originally Posted by SWEngineer
In the "Foundation Maths" book in chapter (19): "The exponential function", there was an example solving the following problem (I used "^" to denote power):

(e^2t + 2e^2 + 1)^1/2

In solving the problem, this was converted to ((e^t + 1)^2)^1/2

What clue can tell that I have to solve the problem this way and not by just taking the root of every element?

Thanks.
The "clue" is recognizing perfect squares. x^2+ 2x+ 1= (x+ 1)^2

Here, x= e^t so x^2= (e^t)^2= e^{2t}

4. Thanks a lot for your replies. It is clear now.