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Mathematics 20 Online
OpenStudy (anonymous):

\[\sqrt{12}+4\sqrt{15}\]

OpenStudy (saifoo.khan):

\[\Huge 4\sqrt{15}+2\sqrt{3}\]

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

how did you get that?

OpenStudy (saifoo.khan):

Calc.

OpenStudy (anonymous):

well yeah, but how did you randomly get a two and a three?

OpenStudy (saifoo.khan):

let me solve

OpenStudy (anonymous):

okay

OpenStudy (saifoo.khan):

\[\huge \sqrt{12} = \sqrt{4\times 3}=2\sqrt{3}\]

OpenStudy (anonymous):

4x3 isnt the sq. root of 12?

OpenStudy (saifoo.khan):

you wont be getting the answer that way.

OpenStudy (anonymous):

this is too confusing

OpenStudy (anonymous):

Jocelyn, are you doing GCSE maths- surds?, If you are I can try and help you with this.

OpenStudy (anonymous):

i dont know what that is

OpenStudy (saifoo.khan):

\[\huge \sqrt{12} = \sqrt{4\times 3}=\sqrt{2 \times 2 \times 3}=2\sqrt{3}\]

OpenStudy (saifoo.khan):

now can you get it?

OpenStudy (anonymous):

Basically, what you want to do is get √12 and find 2 numbers that times together to get 12, for example- saifoo.khan did √4×3 - 4 x 3 = 12. Then once you have that, you must find the square root of a square number, which in this case is 4. Take the 4 and find the square root, which is 2, and thats the 2√3

OpenStudy (anonymous):

okay, the explanation helped a bit.... thanks (:

OpenStudy (anonymous):

so would my answer be \[2\sqrt{3}\]

OpenStudy (saifoo.khan):

no the answer will be my first post.

OpenStudy (anonymous):

4√15 +2√3, saifoo's first answer given.

OpenStudy (anonymous):

okay...

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