#1

Can someone help me figure this problem?

b + 2 = square root of 6b + 3

b + 2 = square root of 6b + 3

#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?

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

#3

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.

#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

There you go, hopefully the process makes sense, so you'll be able to apply the same principles for future problems

(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

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

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

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

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?

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

#5

is it sqrt(6b) + 3 or sqrt(6b + 3) ?

#6

When I first did this, I subtracted the b and 2 to make square root of 5b+1.

I don't math...

I don't math...

#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.

If it's sqrt(6b) + 3, essentially do the same thing, but move the three over before you square both sides.

#8

is it sqrt(6b) + 3 or sqrt(6b + 3) ?

sqrt(6b+3)

#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.

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

#10

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)

#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

Glad it helped

#12

Thanks!

#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).

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.

Nerds, the lot of you.

#15

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

you don't even know

#16

Lol U Said Squirt

#17

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

nah

not even close

TS's problem was quadratic. i could have pulled out original arab geometric solutions to quadratics and done it that wayyou don't even know

#19

omg user of the year right here.

5 stars

then why didn't you vote for me you knobhead

#20

Nerds.

Nerds, the lot of you.

pls garcia, this is basic algebra