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Mathematics 8 Online
OpenStudy (anonymous):

how do you solve sqrt2(5+sqrt8)^2 ? A. 66+40 x sqrt2 B. 40+33 x sqrt2 C. 33 x sqrt2 D. 5 x sqrt2 + 4

OpenStudy (goformit100):

A. 66+40 x sqrt2

OpenStudy (anonymous):

why?

OpenStudy (lgbasallote):

\[\sqrt 2(5 + \sqrt 8)^2\] right?

OpenStudy (zzr0ck3r):

i thought it was sqrt(2(5+sqrt(8))^2)

OpenStudy (anonymous):

yes @igbasa

OpenStudy (lgbasallote):

uhmm okay..do you know the shortcut to square of a binomial? \[(a+b)^2 = a^2 + 2ab _ b^2\]

OpenStudy (lgbasallote):

\[(a+b)^2 = a^2 + 2ab+ b^2\]

OpenStudy (anonymous):

nope :/

OpenStudy (lgbasallote):

uhmm okay we can do it the long way

OpenStudy (lgbasallote):

first let us solve \[(\sqrt 5 + 8)^2\]

OpenStudy (anonymous):

And how do we do that?

OpenStudy (lgbasallote):

\[(\sqrt 5 + 8)^2 \implies (\sqrt 5 + 8)(\sqrt 5+ 8)\] agree?

OpenStudy (anonymous):

yess

OpenStudy (lgbasallote):

you're familiar with FOIL right?

OpenStudy (anonymous):

very much so

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

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

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):

\[8\sqrt{5} +8\sqrt{5} + 69\]

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 \[8x + 8x\]

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: \[1) \; a \times \sqrt b = a\sqrt b\] \[2) \; \sqrt c \times d \sqrt e = d\sqrt{c\times e}\]

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):

\[69\sqrt{2}?\]

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

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?

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

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 :/

OpenStudy (lgbasallote):

remember \[\sqrt a \times b\sqrt c = b\sqrt{a \times c}\] \[\sqrt 2 \times 10\sqrt 8 = 10\sqrt{8 \times 2}\] got it?

OpenStudy (anonymous):

lol yup

OpenStudy (lgbasallote):

so what is 8 x 2?

OpenStudy (anonymous):

16

OpenStudy (lgbasallote):

and what's the square root of 16?

OpenStudy (anonymous):

4

OpenStudy (lgbasallote):

so we have \[\sqrt 2 \times 10 \sqrt 8 \implies 10 \sqrt{8\times 2} \implies 10\sqrt {16} \implies 10 \times 4\] did you get that?

OpenStudy (anonymous):

everything up until 10 x 4

OpenStudy (lgbasallote):

\[10 \sqrt {16} \implies 10 \times \sqrt{16} \implies 10 \times 4}\] now do you get it?

OpenStudy (lgbasallote):

\[10 \sqrt {16} \implies 10 \times \sqrt{16} \implies 10 \times 4\]

OpenStudy (anonymous):

*dummy slaps self*

OpenStudy (lgbasallote):

heh lol. so what do you get now?

OpenStudy (anonymous):

40 + 33 radical2

OpenStudy (anonymous):

:DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD THANK YOU SO MUCH. FOR YOUR PATIENCE AND KINDNESS AND EVERYTHING

OpenStudy (lgbasallote):

hehe <tips imaginary hat> sorry for the constant delayed replies though

OpenStudy (anonymous):

:) It's alright. I can't hog your services hehe. Have a really good night(or day depending where you are right now) Igbas :).

OpenStudy (lgbasallote):

hehe you too ^_^

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