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Square root questions 
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#1
Apr407, 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] 


#2
Apr407, 07:46 PM

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P: 8,305

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.] 


#3
Apr407, 07:51 PM

P: 21

[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) 


#4
Apr407, 08:41 PM

HW Helper
P: 1,542

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. 


#5
Apr407, 09:36 PM

P: 1,345

Also, remember to rationalize the fraction^^



#6
Apr507, 03:30 AM

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P: 8,305

Can you simplify this? 


#7
Apr507, 07:35 AM

P: 1,635

[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?? 


#8
Apr507, 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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