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Mathematics 17 Online
OpenStudy (anonymous):

Help, please?

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 ?

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???

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?

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..

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):

\(\large b^4 = 4^2*25\)

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):

\(\large b^2 = \pm 4*5\) \(\large b = \pm 2 \sqrt{5}\)

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

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

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?

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???????

ganeshie8 (ganeshie8):

Yes its B ! good guess :)

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