Square Root Simplified: Understanding the Result as a Single Nonnegative Number

In summary, the conversation discusses simplifying expressions involving the square root of u^2 / c^2 and how it is equal to v in the end. The process involves canceling out terms and taking only the positive root in the final answer. However, there is a discrepancy between the solution shown in the original post and the one shown in another book.
  • #1
lioric
306
20
relativistic 4.jpg

How is it equal to v in the end?
I'm sorry for asking such questions. But I'm just trying to understand
 
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  • #2
Because addition is associative and commutative and ##c^2≠0##. For short: just calculate the left hand side.
 
  • #3
Can you simplify the denominator?
 
  • #4
DrClaude said:
Can you simplify the denominator?
Dear god I must be blind
v^2-v^2 = 0
c^2 and c^2 cancels
And all that's left is a rooted v^2 which cancels itself

But look at this
relativistic 5.jpg

This is how the same solution is illustrated in another book.
(1-u^2/c^2) cancels each other
The u^2 subtracts each other in the denominator like in the previous sollution
Which leaves a rooted u^2 / c^2
How does that simplify?
 
  • #5
lioric said:
Which leaves a rooted u^2 / c^2
How does that simplify?
Look to the left of the square root sign...
 
  • #6
DrClaude said:
Look to the left of the square root sign...
I can see the c root What does it mean?
 
  • #7
lioric said:
I can see the c root What does it mean?
No, it's c times the root.
 
  • #8
DrClaude said:
No, it's c times the root.
So are you saying that the large root will cancel the squares of u^2 / c^2 making it like u / c * c/1
and c and c cancels?
 
  • #9
lioric said:
So are you saying that the large root will cancel the squares of u^2 / c^2 making it like u / c * c/1
and c and c cancels?
Yes. Note that the author there takes only the positive root, while in the OP the two roots are kept.
 
  • #10
DrClaude said:
Yes. Note that the author there takes only the positive root, while in the OP the two roots are kept.
Thank you very much
 
  • #11
lioric said:
View attachment 96291
How is it equal to v in the end?
I'm sorry for asking such questions. But I'm just trying to understand
This is actually incorrect. On the right side it should be |v|, not ##\pm v##. In other words, the square root evaluates to a single nonnegative number, not two numbers.
 

Related to Square Root Simplified: Understanding the Result as a Single Nonnegative Number

What is a square root and why is it important?

A square root is a mathematical operation that finds the number which, when multiplied by itself, gives a certain number. It is important because it allows us to find the length of a side in a right triangle, to solve equations, and to simplify radical expressions.

How do I simplify a square root?

To simplify a square root, you can factor the number inside the radical and remove any perfect square factors. Then, write the remaining factors as the square root of a whole number. For example, the square root of 18 can be simplified to the square root of 9 times the square root of 2, which becomes 3√2.

Can all square roots be simplified?

No, not all square roots can be simplified. Some square roots, like √5 or √7, are irrational numbers and cannot be simplified to a whole number. They can only be written in decimal form or as an infinite series.

What is the difference between simplifying and rationalizing a square root?

Simplifying a square root involves simplifying the number inside the radical, while rationalizing a square root involves removing any radicals from the denominator of a fraction. Rationalizing is usually done to make expressions easier to work with or to eliminate imaginary numbers.

What are some real-world applications of simplifying square roots?

Simplifying square roots has many real-world applications, such as calculating the height of a tree or building, finding the length of a diagonal in a rectangle, or determining the amount of wire needed to fence a circular garden. It is also used in fields such as engineering, physics, and finance.

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