What's this weird square root ??

[Yuseph]

Hey guys,

Sorry still a beginner im only at chapter 4. Whats the first square root ?? How do you read that ? And why is the result 3 ?

Btw how do you read the second square root. I understand the concept i know its a simplified form i know its 4 * 4 * 81 and all but how do you read it ?

 

[Dr.Peterson]

The expression [MATH]\sqrt[4]{81}[/MATH] means the fourth root of 81, that is, the positive number whose fourth power is 81. And, in fact, [MATH]3^4 = 81[/MATH]. It isn't a square root at all.

The second, [MATH]4\sqrt{81}[/MATH], is 4 times the square root of 81; it's better explained as 4 times 9, which is 36.

 

[Deleted member 4993]

View attachment 20954

Hey guys,

Sorry still a beginner im only at chapter 4. Whats the first square root ?? How do you read that ? And why is the result 3 ?

Btw how do you read the second square root. I understand the concept i know its a simplified form i know its 4 * 4 * 81 and all but how do you read it ?

You asked:

Whats the first square root ?? How do you read that ? And why is the result 3 ? \(\displaystyle \sqrt[4]{81} = ?\)

First of all \(\displaystyle \sqrt[4]{-}\) is not "read" as square -root. It is quartic root or square root of square root.

Concept is very similar to that of square root. What number - when multiplied by itself - 4 times - produces 81. Of course we know that 3*3*3*3 = 81. So:

\(\displaystyle \sqrt[4]{81} \ = \ \sqrt[4]{3*3*3*3} \ = \ 3\)

 

[Otis]

… i know [it's] a simplified form
i know its 4 * 4 * 81 …

Hi Yuseph. Actually, 4√81 is a simpler form than √1296.

If they'd changed 4√81 to √1296, as your post suggests, then they've taken the factor 4 (outside the radical sign) and moved it inside the radical sign as 4².

4√81 = √16 · √81 = √(16·81) = √1296

It's not necessary to do that, if the task is to evaluate 4√81.

4√81 = 4(9) = 36

?

 

[JeffM]

[MATH]4 * \sqrt{81} = 4 * 9 = 36.[/MATH]
The root sign with vinculum is both an operator and a grouping symbol. In my opinion,

[MATH]4 \sqrt{81} = \sqrt{16 * 81} = \sqrt{1296} = 36[/MATH]
is a technical violation of PEMDAS as well as being unnecessarily complicated.

 

[Dr.Peterson]

In my opinion,

[MATH]4 \sqrt{81} = \sqrt{16 * 81} = \sqrt{1296} = 36[/MATH]
is a technical violation of PEMDAS as well as being unnecessarily complicated.

I wouldn't say it violates PEMDAS; that's not a rule that you can only carry out the evaluation in that order, just a rule that interprets the meaning of an expression. For example, you can choose to add a string of positive numbers and negative numbers separately, rather than left to right. More to the point, you are allowed to distribute before you evaluate (though it's usually a waste of time), and this is a lot like distributing (and yes, it's a waste of time, as well as confusing students who read it).

 

[HallsofIvy]

It very greatly annoys me that, in order to write, say, the fourth root in Latex, \(\displaystyle \sqrt[4]{x}\), you have to write "\sqrt[4]{x}".