./(6x+1) + ./(x-4) = 5

./(-) = Square root of

Any help would be appreciated.

I have tried many times and cannot seem to come up with a correct answer.
I know 100% that there must be a correct answer, so if an explanation was given to me on how to come about solving this equation it would help me tremendously!

looks like x=4?

./(6x+1) + ./(x-4) = 5 = f(x)

since the domain of sqrt is [0, inf), we know x - 4 >= 0, so x >= 4. that's the first one i tried which worked. so i got lucky. if you increase x, f(x) increases. i was going to try integers since f(x) is an integer.

dunno how to solve it algebraically, sorry. or rather, i dont care enough to try lol.

This post was edited by carteblanche on Sep 28 2014 02:50pm

Yes, I did get that answer too.

But these are the possible answers.

A. (4,8)
B. (-4,8)
C. (-4,-8)
D. (4,-8)
E. No Solution (However, I know this cannot be the correct answer)

I plug in 4 to the equation and it works, but none of the others come out to be true statements if I plug them in.

Correct me if I am wrong.



haha, thanks for your help though!

Still looking for an explanation of how to solve algebraically from someone please =]

This post was edited by Iffy00 on Sep 28 2014 02:56pm

well, technically the square root can be positive or negative. so 8 could work, i think? +sqrt(49) + -sqrt(4) = 5. given the choices, i'd pick A. x certainly cannot be less than 4, so it's def not B, C, D

This post was edited by carteblanche on Sep 28 2014 02:56pm

Yes, but 7+2 =/= 5

So 8 cannot be correct... This was a test question, I picked no correct solution but got it wrong.

I'm just trying to comprehend how to come up with the correct solution and then check for if it is correct.

idk man...i'm done with school

just go with process of elimination. you know there is a solution, so it's not E. and x can't be negatives so it's not B, C, D. in the words of sherlock holmes, when you have eliminated the impossible, whatever remains, however improbable, must be the truth

lucky lol, I just want to be able I know how to solve such equations in case I see a similar in the future.

There isn't an algebraic way to solve many seemingly simple equations
What math class are you in?

College Algebra lol...
There has to be a way to solve it, for example single out the radical then ^2 both sides.
The problem is simplifying to where I get x= one of these answers.

4 has been the only number to work but that is not the only answer.

/e Logic says that A must be the correct answer, but when I check my answers 8 does not work.

So is this a trick question? if so, why was I not correct when answering no solution.
If A is the correct answer, and 8 does not work, it must be designed to trick me somehow.. i'm just trying to figure out.

I had a similar question to where I got the answer and none of the answers had parenthesis, but the numbers that were listed were correct.
When I answered with the numbers I got, the answer was wrong, so because there were no parenthesis none of them were the real solution.

My teacher is either really humorous and does this on purpose, or he accidentally designed questions wrong.

This post was edited by Iffy00 on Sep 28 2014 04:41pm

8 is a possible solution, so it does work, if you break down that equation into 4 equations using the positive/negative sqrts

similarly, if you say x = sqrt(4) + 1, you have two solutions: 2 + 1 =3 and -2 + 1 = -1

This post was edited by carteblanche on Sep 28 2014 04:44pm