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Square root questions

by w3tw1lly
Tags: root, square
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w3tw1lly
#1
Apr4-07, 07:41 PM
P: 21
I feel embarassed to ask these questions but what is the rule for to simplify division, addition, and subtraction square roots? Here are some questions:

SIMPLIFY:


[tex]5\sqrt{24}\div2\sqrt{18}[/tex]



[tex]\sqrt{40} + \sqrt{90}[/tex]




[tex]\sqrt{50} - \sqrt{18}[/tex]
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cristo
#2
Apr4-07, 07:46 PM
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What have you tried? You need to simplify the square roots. For example, write the first question as [tex]\frac{5\sqrt{24}}{2\sqrt{18}}[/tex] Now, can you simplify [itex]\sqrt{24}[/itex] and [itex]\sqrt{18}[/itex]?

[Hint: write each number under the sqrt sign as a product of primes.]
w3tw1lly
#3
Apr4-07, 07:51 PM
P: 21
Quote Quote by cristo View Post
What have you tried? You need to simplify the square roots. For example, write the first question as [tex]\frac{5\sqrt{24}}{2\sqrt{18}}[/tex] Now, can you simplify [itex]\sqrt{24}[/itex] and [itex]\sqrt{18}[/itex]?

[Hint: write each number under the sqrt sign as a product of primes.]
Sorry, I meant to write the question like a fraction I just didn't know the code. When you are simplifying roots, and you take out lets say the root of 4, do you times the number already outside the root sign by 2?
[tex]\frac{5\sqrt{24}}{2\sqrt{18}}[/tex]
=[tex]\frac{5\sqrt{4*6}}{2\sqrt{3*6}}[/tex] (don't know what to do, so long since we had done radicals)

hage567
#4
Apr4-07, 08:41 PM
HW Helper
P: 1,540
Square root questions

"When you are simplifying roots, and you take out lets say the root of 4, do you times the number already outside the root sign by 2?"

Yes.
Feldoh
#5
Apr4-07, 09:36 PM
P: 1,345
Also, remember to rationalize the fraction^^
cristo
#6
Apr5-07, 03:30 AM
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P: 8,309
Quote Quote by w3tw1lly View Post
Sorry, I meant to write the question like a fraction I just didn't know the code. When you are simplifying roots, and you take out lets say the root of 4, do you times the number already outside the root sign by 2?
[tex]\frac{5\sqrt{24}}{2\sqrt{18}}[/tex]
=[tex]\frac{5\sqrt{4*6}}{2\sqrt{3*6}}[/tex] (don't know what to do, so long since we had done radicals)
[tex]\frac{5\sqrt{4*6}}{2\sqrt{3*6}}=\frac{5\cdot 2\cdot\sqrt{6}}{2\cdot\sqrt{3}\cdot\sqrt{6}}[/tex]

Can you simplify this?
sutupidmath
#7
Apr5-07, 07:35 AM
P: 1,635
Quote Quote by w3tw1lly View Post
I feel embarassed to ask these questions but what is the rule for to simplify division, addition, and subtraction square roots? Here are some questions:







[tex]\sqrt{40} + \sqrt{90}[/tex]




[tex]\sqrt{50} - \sqrt{18}[/tex]

[tex]\sqrt{40} + \sqrt{90}[/tex]=[tex]\sqrt{4*10}+\sqrt{9*10}[/tex]=2[tex]\sqrt{10}+3\sqrt{10}[/tex]=

can you go from here??
thomas49th
#8
Apr5-07, 08:48 AM
P: 656
this may confuse you more but when you add fractions you need to get the denominator (number on the bottom of fraction) the same. The same goes with surds (square roots), you need to get the number inside the root the same on each surd in oder to add/subtract.


I find it harder to do the + - surds than the x and / surds

When you divide:
[tex] \sqrt{a} \div \sqrt{b} = \frac {\sqrt{a}}{\sqrt{b}}[/tex] which is also written as [tex]\sqrt{\frac{a}{b}}[/tex]

Have a look

http://www.mathsrevision.net/gcse/pages.php?page=6

and

http://www.bbc.co.uk/schools/gcsebit...rdshrev2.shtml


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