Is There a Method to Find the Square Root of Confusing Algebraic Expressions?

[Miike012]

I found a way to find the square root of an expression which is confusing to me.
Is anyone familiar to this method? And my main question... will it be useful?

 

[eumyang]

I am not familiar with this method. I found a http://www.youtube.com/watch?v=iz5c0DizXyk" that shows how to do a similar problem. It's not in English, though (Hindi?). But between the video and your attachment I was able to figure out how it works. As to whether or not this is useful, I personally don't find it useful, but others may have a different opinion.

 

[Mark44]

I don't think this is very useful, either, and certainly not worth memorizing. What is much more useful is being able to factor perfect square trinomials such as x² + 4x + 4 = (x + 2)² and the like.

 

[Miike012]

By looking at the expression I would't have even guess it was a perfect square.. I thought it would have been something along the lines of...

(cx + a)^4 ... with some type of coefficient infront of x.

 

[QuarkCharmer]

If you memorize (or can derive) a few lines of "Pascals Triangle", you can quickly figure out things like (x+y)^5 and so on. I found that worth learning.

 

[eumyang]

By looking at the expression I would't have even guess it was a perfect square.. I thought it would have been something along the lines of...

(cx + a)^4 ... with some type of coefficient infront of x.

Well, if the original polynomial ended up being a binomial raised to the 4th power, then the original polynomial would still be a perfect square, would it not? Using your notation,
(cx + a)⁴ = [((cx + a)²]², after all.

But as it is, the original polynomial is not a binomial raised to the 4th power. If you rearrange the terms, the last one, 4a⁴, is NOT a perfect 4th-power. In other words, you can't write 4a⁴ as (ka)⁴, where k is an integer.