#1
Can someone help me figure this problem?

b + 2 = square root of 6b + 3
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#2
Ok I'm going to help you through this by steps, so that you actually figure out how to do the math, rather than how to get the answer.

First thing, what is the biggest 'obstacle' in finding the solution?
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#3
Quote by Baby Joel
Ok I'm going to help you through this by steps, so that you actually figure out how to do the math, rather than how to get the answer.

First thing, what is the biggest 'obstacle' in finding the solution?


I would say the square root.
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#4
right. So what you'd need to do is figure out how to get rid of the square root

(I'll put everything in spoilers from here on out, just because it's easier that way)
In order to get rid of the square root, you have to square both sides
This gives you (b+2)^2=6b+3
that ^ means to the power of, just so you are aware
Now the problem is much easier to solve
Rewritten it becomes (b+2)(b+2)=6b+3
So just multiply the left side out
b^2+4b+4=6b+3
At this point I like to move everything over to one side
Subtract the right from both sides, meaning you work out b^2+4b+4-(6b+3)=6b+3-(6b+3)
This gets you b^2-2b+1=0
Now it's just a normal binomial or whatever the term is, so factor it out
you get (b-1)(b-1)=0
So what value of 'b' will give you zero when you subtract one from it?
b=1


There you go, hopefully the process makes sense, so you'll be able to apply the same principles for future problems
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#6
When I first did this, I subtracted the b and 2 to make square root of 5b+1.

I don't math...
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#7
Oh good catch. TS, in my solution, I worked it out for sqrt(6b+3)
If it's sqrt(6b) + 3, essentially do the same thing, but move the three over before you square both sides.
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#8
Quote by Eastwinn
is it sqrt(6b) + 3 or sqrt(6b + 3) ?

sqrt(6b+3)
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#9
In that case, just work it out using the steps I gave through the spoilers in my earlier post.

If you have questions about a specific part, wondering why to do that certain step, feel free to ask.
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#10
Quote by Baby Joel
In that case, just work it out using the steps I gave through the spoilers in my earlier post.

If you have questions about a specific part, wondering why to do that certain step, feel free to ask.


THANK YOU. I figured 5 problems using that. How would I work that concept in a problem like this? -4 = -v + sqrt(13-4v)
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my last fail was breaking up with my gf.

that's going to suck for a while


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#11
It's the same idea. To get rid of a square root, isolate it on one side. So with that one, move the -v to the left side, so you get v-4=sqrt(13-4v) and then do the same process of squaring both sides, etc.

Glad it helped
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#12
Thanks!
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my last fail was breaking up with my gf.

that's going to suck for a while


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Well, not really haha!


#13
Also, remember that the square-rooted bit must be non negative. In that last one, for instance, you have to have 13 - 4v ≥ 0 which means v ≤ 13/4. If any of the solutions you get in the end is larger than 13/4 then you have to exclude it.

Or even easier, check the value of 13 - 4v for your solution(s).
Last edited by sickman411 at Sep 2, 2014,
#14
Nerds.

Nerds, the lot of you.
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#15
Quote by Joshua Garcia
Nerds.

Nerds, the lot of you.


nah

not even close

TS's problem was quadratic. i could have pulled out original arab geometric solutions to quadratics and done it that way

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#17
Quote by Baby Joel
It's the same idea. To get rid of a square root, isolate it on one side. So with that one, move the -v to the left side, so you get v-4=sqrt(13-4v) and then do the same process of squaring both sides, etc.

Glad it helped

omg user of the year right here.


5 stars
#18
Quote by Eastwinn
nah

not even close

TS's problem was quadratic. i could have pulled out original arab geometric solutions to quadratics and done it that way

you don't even know



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#19
Quote by Ssargentslayer
omg user of the year right here.


5 stars

then why didn't you vote for me you knobhead
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#20
Quote by Joshua Garcia
Nerds.

Nerds, the lot of you.

pls garcia, this is basic algebra