Simplifying √(1)2+(2t)2+(2t2)2= 1+2t2: Understanding the Solution

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The discussion revolves around simplifying the expression √(1)² + (2t)² + (2t²)² = 1 + 2t². Participants clarify that the correct simplification leads to √(1 + 4t² + 4t⁴), which cannot be canceled due to the presence of the 1 term. They emphasize that the left-hand side should be simplified to match the right-hand side without any cancellations. One participant notes that recognizing the form a² + 2ab + b² is key to understanding the simplification process. Overall, the conversation highlights common algebraic misunderstandings and the importance of practice in mastering such concepts.
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Homework Statement



√(1)2+(2t)2+(2t2)2= 1+2t2

I'm a bit confused I'm getting √(1+4t^2+4t^2) don't quite understand how they are getting the above answer.

Thanks


Homework Equations





The Attempt at a Solution

 
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Mdhiggenz said:

Homework Statement



√(1)2+(2t)2+(2t2)2= 1+2t2

I'm a bit confused I'm getting √(1+4t^2+4t^2) don't quite understand how they are getting the above answer.

Thanks
An algebra mistake:##\sqrt{1+(2t)^2+(2t^2)^2}=\sqrt{1+4t^2+4t^4}## Does that help?
 
No, that was a typo on my part. I originally got the same thing as you. If you factor out a 4t^2

You get √1+4t^2(1+t^2). I don't see how anything can be canceled due to that pesky 1 term.
 
Mdhiggenz said:
No, that was a typo on my part. I originally got the same thing as you. If you factor out a 4t^2

You get √1+4t^2(1+t^2). I don't see how anything can be canceled due to that pesky 1 term.

There's no cancellation here. You're merely simplifying the left hand side to get the right hand side.

What is {(1 + 2t^2)}^2?
 
Haha wow silly me. This has always actually confused me, is there a set formula to condense it the way you did?

Thanks again for the help
 
Mdhiggenz said:
Haha wow silly me. This has always actually confused me, is there a set formula to condense it the way you did?

Thanks again for the help

No, just observation of the form a^2 + 2ab + b^2. I guess it comes with experience and practice. :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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