Simplifying square roots?

2008-01-08 7:03 am
How do you do these:

1. 4√6*√6
2. 5√24*3√10
3. 8/√3
4. 2/1-√3

回答 (7)

2008-01-12 5:48 am
✔ 最佳答案
1. = 4 * 6 = 24
2. =5* √(4* 2 * 3) *3√(2 * 5) = 10 * 3√(2*2*3*5 ) = 30 * 2√3*5 = 60√15
3. 8/√3 , multiply √3 on both denominator and enumerator, = 8√3 /3
4. 2/1-√3, multiply 1 +√3 on both denominator and enumerator, = 2* (1+√3) /( 1-3)= 2* (1+√3) /( -2) = (1+√3) /( -1) = -1 -√3)
2008-01-08 10:51 pm
Question 1
4√6 x √6 = 4 x 6 = 24

Question 2
15 √24 √10
30 √6 √10
30√ 60
60 √15

Question 3
8 / √3
8√3 / 3

Question 4
2(1 + √3) / (-2)
- (1 + √3)
2008-01-08 3:28 pm
1. 4√6*√6 = 4√(6*6)
= 4√36
= 4*6
= 24
2. 5√24*3√10 = 5*3√(24*10)
= 15√240
= 15*15.49
= 232.38
3. 8/√3 = (8/√3) * (√3/√3)
= 8√3/√(3*3)
= (8√3)/3 OR 4.62
4. 2/1-√3 = (2/1-√3) * (1-√3/1-√3)
= [2(1-√3)] / [(1-√3)(1-√3)]
= (2-2√3) / 1-2√3-√(3*3)
= (2-2√3) / 1-2√3-3
OR
= -1.46/-5.46
= 0.27
參考: note: i'm not sure with all the calculations.... c",)
2008-01-08 3:26 pm
1. notice that you have two identical √6.. and so when you square them (multiply two of the same things) you get 6! so the answer for 1 is just 4*6 = 24

2. multiply the whole numbers, to get 15, then multiply the numbers under the √ to get √240: 15√240

then you can factor out the √240 into √2*2*2*2*3*5

pair off what you can, and move it outside (so √2*2 = 2, and √2*2 = 2) 2*2*15*√3*5

60√15 is as simple as it gets!

3. the idea here is just to get the √3 out from the bottom:

√3 * 8..........8√3
---------- =....----------
√3 * √3........3

4. again, gotta get the √3 off the bottom, so multiply top & bottom by 1 + √3 (notice you flipped the - to a +... this makes the √3 go away! this is a little trick you just kinda have to know...)

(1 + √3) * 2
---------------
(1 + √3) * (1 - √3)

2 + 2√3
-------------
1 - 3


2 + 2√3
------------
... -2
2008-01-12 2:10 pm
1. = 4 * √(6 ^ 2) = 4 * 6 = 24

2. =5* √(4* 2 * 3) *3√(2 * 5) = 10 * 3√(2*2*3*5 ) = 30 * 2√3*5 = 60√15

3. You need to get rid of sqrt in denominator.
8 * √3 /√3 *√3 = 8√3 /(√3 ^ 2) = 8√3 /3

4. multiply 1 +√3 on both denominator and enumerator,

= 2* (1+√3) /( 1-3)= 2* (1+√3) /( -2) = (1+√3) /( -1) = -1 -√3
2008-01-08 3:16 pm
Use the fact that sqrt(x)*sqrt(y)=sqrt(xy). (This fact follows from the fact that both sqrt(x)*sqrt(y) and sqrt(xy) both have the same square: xy, and both are positive, so they're equal.)

That fact, together with factoring the "radicands" (numbers inside the square root signs) into primes should do it for you.
2008-01-08 3:16 pm
4√6*√6 = (4*1)√(6*6) = 4√36 = 4(6) = 24

8/√3 = 8/√3 * √3/√3 = (8√3)/3 = (8/3)√3


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