Question

anyone here good at adding radical expressions?

Answer 1

give a problem

Answer 2

\[9\sqrt{2x^3}+3\sqrt{18x}\]

Answer 3

oh! the 3 is 3x sorry ^^:

Answer 4

ok: so what you need to add radicals is the same thing under both.
so starting with: \[3x \sqrt{18x}\] 18 can be broken into 9 * 2 and you can take the 9 out of the square root so it becomes: \[3*3\sqrt{2x} = 9\sqrt{2x}\]
next on the left: \[x ^{3} = x ^{2}* x\] so the x^2 can be taken out and put on outside leaving this. \[9x \sqrt{2x}+9x \sqrt{2x} = 18x \sqrt{2x}\]

Answer 5

why do you take the 9 out of square root and make it look that way? sorry for the dumb question, i have issues grasping things.

Answer 6

well 18 is factored by 9*2, 9 is a perfect square. meaning the square root of 9 is an integer. 3*3 = 9 and the square root of 9 is 3

Answer 7

in order to add 2 square roots together the term under the radical needs to be identical

Answer 8

so the goal is to make the ones in square roots to be identical then solve?

Answer 9

when adding up terms in an expression, you can only collect together those terms that have the same type of variable.
to take a simple example, lets say you had:
2x+4y+x-3y
here you can collect the "2x" and "x" to make "3x"
and you can collect the "4y" and "-3y" to make "y"
resulting in:
3x+y

the same principal holds if you have terms that contains powers, e.g. in:\[2x^2+4x+x^2-3x\]you can collect together the terms involving "x^2" and separately the terms involving "x" to get:\[3x^2+x\]
do you understand so far?

Answer 10

yes

Answer 11

ok, now the power terms can be any power - including square roots, so:\[2\sqrt{x}+4x+\sqrt{x}-3x=3\sqrt{x}+x\]

Answer 12

the next thing you need to understand is how to simplify a term

Answer 13

you need to know the following rules:\[a^n.a^m=a^{n+m}\]\[(a^n)^m=a^{nm}\]\[(ab)^n=a^n.b^n\]
are you familiar with these rules?

Answer 14

i think ive seen em before, but I still dont understand them that well

Answer 15

ok, lets take some simple examples

Answer 16

\[2^3=2*2*2\]\[2^2=2*2\]so\[(2^3)*(2^2)=(2*2*2)*(2*2)=2*2*2*2*2=2^5\]this shows the rule:\[a^n.a^m=a^{n+m}\]understand?

Answer 17

i think so

Answer 18

what are you not sure of?

Answer 19

can you help me solve a problem step by step?

Answer 20

sure, but you need to understand these basic things first

Answer 21

\[2\sqrt{27}-2\sqrt{18}+\sqrt{75}\]

Answer 22

try and explain with this problem, please? :) If you dont mind of course.

Answer 23

ok - lets try :-)

Answer 24

remember one of the rules above:\[(ab)^n=a^n*b^n\]this applies to ANY power, so square roots for example are just power 1/2:\[(ab)^{1/2}=a^{1/2}*b^{1/2}\]in other words:\[\sqrt{ab}=\sqrt{a}*\sqrt{b}\]we can use this here:\[2\sqrt{27}=2\sqrt{9*3}=2\sqrt{9}*\sqrt{3}=2*3*\sqrt{3}=6\sqrt{3}\]
do you see how that was achieved?

Answer 25

yes

Answer 26

ok, now try and do the next term yourself - how can you simplify:\[2\sqrt{18}\]

Answer 27

2*6\[\sqrt{3}\]

Answer 28

no quite - try using the same steps that I used - try again please

Answer 29

ok jus a sec..

Answer 30

\[2*2\sqrt{3} \] ? do i take the second three out?

Answer 31

oh wait is it like 3*3*2, and you couple the threes and bring out a 3??

Answer 32

yes

Answer 33

so it would be \[2*3\sqrt{2}\] ?

Answer 34

perfect!

Answer 35

:D really? yay!!

Answer 36

so basically what you need to do is look at each square root and see if you can split the number which is inside the square root into a product of two numbers, one of which is a square number. so the one you did would go something like this:\[2\sqrt{18}=2\sqrt{9*2}=2\sqrt{9}\sqrt{2}=2*3\sqrt{2}=6\sqrt{2}\]

Answer 37

now try the same on the remaining term

Answer 38

5*5*3? Does that look right?

Answer 39

yes 75=5*5*3

Answer 40

but the middle term doesnt share the same number in square root as the first and last? What do i do with it?

Answer 41

\[its a \sqrt{2} and the others are \sqrt{3}\]

Answer 42

thats fine, it just means that these terms cannot be "collected" together, e.g.:\[3\sqrt{2}+4\sqrt{3}+5\sqrt{2}=8\sqrt{2}+4\sqrt{3}\]no more simplification is possible here

Answer 43

so it will just be part of final answer?

Answer 44

yes

Answer 45

\[2*3\sqrt{3}-2*3\sqrt{2}+5\sqrt{3} \] How do we solve the rest?

Answer 46

your result can be simplified further as follows:\[2*3\sqrt{3}-2*3\sqrt{2}+5\sqrt{3}=6\sqrt{3}-6\sqrt{2}+5\sqrt{3}=11\sqrt{3}-6\sqrt{2}\]

Answer 47

I see it, thanks! :D

Answer 48

by the way - you should not use the term "solve" here. what you are doing here is simplifying a radical expression. "solve" is used when you have some unknown quantities and are asked to find their values, e.g. solve:\[2x=6\]"solution" would be:\[x=3\]

Answer 49

oh, but why is there only one squared 3?

Answer 50

what do you mean?

Answer 51

\[\sqrt{3}\] there is only one

Answer 52

what were you expecting?

Answer 53

oh and where did the six come from?

Answer 54

in this term:\[2*3\sqrt{3}\]you can replace "2*3" with "6" to get:\[6\sqrt{3}\]imagine instead you saw:\[2*3x\]this is just the same as:\[6x\]

Answer 55

i know, i got that :) i just mean the 11 and six in final answer

Answer 56

\[2*3\sqrt{3}-2*3\sqrt{2}+5\sqrt{3}\]\[=6\sqrt{3}-6\sqrt{2}+5\sqrt{3}\]\[=11\sqrt{3}-6\sqrt{2}\]I just collected the terms involving square root of 3. similar example would be:\[6x-6y+5x=11x-6y\]

Answer 57

where did the five go?

Answer 58

oh!
nevermind duh! i see it, sorry ^^;

Answer 59

finally! :-)
I'm glad you now understand how to do these types of problems.

Answer 60

:) Thanks so so much!