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OpenStudy (lgbasallote):
so try foiling that one.
OpenStudy (anonymous):
using the radical o.O . Does it still work the same way?
OpenStudy (lgbasallote):
yup..you just have to know a few things
\[\large 1) \; \sqrt a \times \sqrt a = a\]
\[\large 2) \; c \times \sqrt a = c\sqrt a\]
OpenStudy (anonymous):
I don't understand why the sqrt of something times the sqrt of itself = a single #?
OpenStudy (lgbasallote):
\[\large \sqrt a \times \sqrt a = \sqrt{a \times a} = \sqrt{a^2} = (a^2)^{1/2} \implies a^{2/2} \implies a^1 \implies a\]
i dont know if you're familiar with the steps i just made...but that's how it's done
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OpenStudy (anonymous):
I understand it
OpenStudy (lgbasallote):
oh that's nice then
OpenStudy (anonymous):
so just do foil and leave the answer within the radical ?
OpenStudy (lgbasallote):
yup
OpenStudy (lgbasallote):
but dont forget..that wont be the final answer yet.. you still have to multiply by \(\sqrt 2\) later on
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OpenStudy (anonymous):
\[8\sqrt{5} +8\sqrt{5} + \sqrt{25} + 64 ?\]
OpenStudy (lgbasallote):
what's the sqrt of 25?
OpenStudy (anonymous):
5
OpenStudy (lgbasallote):
right..so \[8\sqrt 5 + 8\sqrt 5 + 5 + 64\]
you can combine 5 and 64
OpenStudy (anonymous):
\[8\sqrt{5} +8\sqrt{5} + 69\]
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OpenStudy (lgbasallote):
right.. now think of \(\sqrt 5\) as x..you can combine 8x + 8x right?
OpenStudy (anonymous):
yes 16x
OpenStudy (lgbasallote):
so what's \[8 \sqrt 5 + 8\sqrt 5\]
OpenStudy (anonymous):
\[8\sqrt{10}?\]
OpenStudy (lgbasallote):
hmm like i said think of \(\sqrt 5\) as x think of \[8 \sqrt 5 + 8\sqrt 5\]
as \[8x + 8x\]
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OpenStudy (anonymous):
I don't quite understand. Does that mean I would add both numbers by itself twice? or just the number outside of the radical
OpenStudy (lgbasallote):
you add the numbers OUTSIDE the radical..the radical stays the same
OpenStudy (anonymous):
\[16\sqrt{5}\]
OpenStudy (lgbasallote):
right so you have \[69 + 16\sqrt 5\]
but this is just \[(\sqrt 5+8)^2\]
you need to multiply by \(\sqrt 2\)
\[\sqrt 2(69 + 16\sqrt 5\]
OpenStudy (lgbasallote):
what you need to sole this is:
\[1) \; a \times \sqrt b = a\sqrt b\]
\[2) \; \sqrt c \times d \sqrt e = d\sqrt{c\times e}\]
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OpenStudy (anonymous):
so what i assume is that radicalC x D.Radical.E = 16Radical10 ?
OpenStudy (lgbasallote):
right!
OpenStudy (anonymous):
:) but what happens to the 69? why wasnt it multiplied by radical 2?
OpenStudy (lgbasallote):
look at number 1 hehe
OpenStudy (anonymous):
\[69\sqrt{2}?\]
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OpenStudy (lgbasallote):
yup
OpenStudy (lgbasallote):
so what's the final answer
OpenStudy (anonymous):
\[69\sqrt{2} + 16\sqrt{10}\]
OpenStudy (lgbasallote):
yup there you go
OpenStudy (anonymous):
but that isn't an answer choice O.o
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OpenStudy (lgbasallote):
what's the answer coice?
OpenStudy (anonymous):
\[A. 66+40\sqrt{2} B. 40+33\sqrt{2} C. 33\sqrt{2} D. 5\sqrt{2}+4\]
OpenStudy (lgbasallote):
WAHHH *facepalm*
OpenStudy (lgbasallote):
we were solving the wrong one T_T you didnt tell me sooner lol
OpenStudy (anonymous):
you're joking right?
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OpenStudy (lgbasallote):
we were solving \[\sqrt 2(\sqrt 5 + 8)^2\]
we should've solved \[\sqrt 2(5 + \sqrt 8)^2\]
sorry
OpenStudy (lgbasallote):
i totally messed that one up :/
OpenStudy (anonymous):
hehe that was funny :3. Hey at least I learned some thing lol :). Maybe this time we can move faster because i have somewhat a clue what i am doing
OpenStudy (lgbasallote):
haha yup
OpenStudy (lgbasallote):
so \[(5 + \sqrt 8)^2 \implies (5 + \sqrt 8)(5+\sqrt 8)\]
now use FOIL
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OpenStudy (anonymous):
\[5\sqrt{8} + 5\sqrt{8} + 33\]
OpenStudy (lgbasallote):
right. now ad \[5\sqrt 8 + 5\sqrt 8\]
OpenStudy (anonymous):
\[10\sqrt{8}\]
OpenStudy (lgbasallote):
right so we have \[\sqrt 2(10 \sqrt 8 + 33)\]
OpenStudy (anonymous):
yes. Now I don't know how to distribute that :/
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