Mathematics
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OpenStudy (anonymous):
Help, please?
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OpenStudy (anonymous):
OpenStudy (anonymous):
@campbell_st @ganeshie8 @.Sam. ?
ganeshie8 (ganeshie8):
for 6 :
start by squaring both sides
ganeshie8 (ganeshie8):
\(\large \sqrt{2x+3} = \sqrt{6x-1} \)
square both sides
\(\large (\sqrt{2x+3})^2 = (\sqrt{6x-1})^2 \)
ganeshie8 (ganeshie8):
any ideas what we can do next ?
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OpenStudy (anonymous):
@ganeshie8 simplify?
ganeshie8 (ganeshie8):
yes :), here is what we can do :- we can cancel square and squareroot.
ganeshie8 (ganeshie8):
that gives -
\(\large 2x+3 = 6x-1 \)
ganeshie8 (ganeshie8):
you can solve x now ?
OpenStudy (anonymous):
@ganeshie8 is it C???
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ganeshie8 (ganeshie8):
No
OpenStudy (anonymous):
D: so sorry, I'm horrible at this~ @ganeshie8
ganeshie8 (ganeshie8):
\(\large 2x+3 = 6x-1 \)
-2x -2x
\(\large 3 = 4x-1\)
+1 +1
\(\large 4 = 4x\)
ganeshie8 (ganeshie8):
its okay :) you wil get better as u do more problems :)
see if u can find x now ?
OpenStudy (anonymous):
@ganeshie8 1?
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ganeshie8 (ganeshie8):
Yes ! good work!
OpenStudy (anonymous):
@ganeshie8 ahh, lifesaver!! :) can you help me with the next one too?
ganeshie8 (ganeshie8):
ah ty :) let me see
OpenStudy (anonymous):
@ganeshie8 wait, is it B?
ganeshie8 (ganeshie8):
same logic, start by squaring both sides..
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ganeshie8 (ganeshie8):
you're almost there, try again :)
ganeshie8 (ganeshie8):
\(\large \sqrt{\frac{b^4}{25}} = 4\)
square both sides
\(\large (\sqrt{\frac{b^4}{25}})^2 = (4)^2\)
ganeshie8 (ganeshie8):
as usual, on left side, the square and square root cancel out
ganeshie8 (ganeshie8):
\(\large \frac{b^4}{25} = 4^2\)
ganeshie8 (ganeshie8):
\(\large b^4 = 4^2*25\)
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ganeshie8 (ganeshie8):
\(\large b^2 = \pm 4*5\)
ganeshie8 (ganeshie8):
\(\large b = \pm ?\)
OpenStudy (anonymous):
20? @ganeshie8
ganeshie8 (ganeshie8):
we need to take square root to get b
ganeshie8 (ganeshie8):
\(\large b^2 = \pm 4*5\)
\(\large b = \pm 2 \sqrt{5}\)
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OpenStudy (anonymous):
@ganeshie8 ty!! 1 more??
ganeshie8 (ganeshie8):
sure, post it as a new quesiton.. il see if i can help :)
OpenStudy (anonymous):
OpenStudy (anonymous):
OpenStudy (anonymous):
@ganeshie8
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ganeshie8 (ganeshie8):
easy, look at the given function
ganeshie8 (ganeshie8):
\(\large y = \sqrt{2x+6}\)
ganeshie8 (ganeshie8):
if u let y = 0,
then,
x = ?
ganeshie8 (ganeshie8):
0 = 2x + 6
x = -3, right ?
ganeshie8 (ganeshie8):
that means, x goes negative, when y goes 0. look at the graph and tell me, in which graph its like that
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OpenStudy (anonymous):
@ganeshie8 C?
OpenStudy (anonymous):
@ganeshie8 ....
ganeshie8 (ganeshie8):
no, C has x value of 4, when y =0
ganeshie8 (ganeshie8):
try again :)
OpenStudy (anonymous):
@ganeshie8 ooohhh C?
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OpenStudy (anonymous):
@ganeshie8
ganeshie8 (ganeshie8):
C is wrong
ganeshie8 (ganeshie8):
look at where that redline is touching he x axis... .
OpenStudy (anonymous):
@ganeshie8 WAIT-B?
OpenStudy (anonymous):
@ganeshie8 IS IT b???????
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ganeshie8 (ganeshie8):
Yes its B ! good guess :)