# Math Help - NEED Help Quick

1. ## NEED Help Quick

Can Someone help me solve these two problems

Absolute value of 2x+1 = x-1 (two solutions i think?)

and

square root of 3x+1 = 3 + square root of x-4

2. Originally Posted by helpless13
Can Someone help me solve these two problems

Absolute value of 2x+1 = x-1 (two solutions i think?)

and

square root of 3x+1 = 3 + square root of x-4

In order to solve an absolute value problem, you have to solve two separate equations. The first will be $2x + 1 = x - 1$ and the second will be $-2x - 1 = x - 1$ or equivalently, $2x + 1 = -x + 1$.

For the square root problem you begin with $\sqrt{3x + 1} = 3 + \sqrt{x-4}$. What you need to do is square both sides, then put everything that's not under a square root on one side, and square both sides again.

3. ## To Iceman

I understand the way your telling me to do it but for the first problem i get 0, -2 which i don't know if they work

and for the second one i get stuck at 2x-4 = 6 times square root x-4 and i dont know how to go anyfurther

For the second problem, you're on the right track. Square both sides of the equation and then put it into the form $ax^2 + bx + c = 0$.

5. ## To Iceman

Okay i got 8 and 5 for the second which check out but 0 and -2 don't check with the original first problem

6. Originally Posted by helpless13
Okay i got 8 and 5 for the second which check out but 0 and -2 don't check with the original first problem
Actually, you're right. The first problem doesn't have any solutions. You have to check, once you get possible answers, whether they really solve the equation. If the problem had been instead $|2x + 1| = |x-1|$ then those two solutions would be correct.

7. ## To Iceman

thanks i really appreciate your help

8. Originally Posted by helpless13
Can Someone help me solve these two problems

Absolute value of 2x+1 = x-1 (two solutions i think?)

and

square root of 3x+1 = 3 + square root of x-4

$|2x+1| = x-1$

You can pull the one out since it is always positive:

$|2x|+1 = x-1$

$|2x|-x = -2$

From here it should be clear that, as you have already discovered, there are no solutions, because no matter the value of x, |2x| will always be greater than it, and so |2x|-x will always be greater than or equal to zero, and thus can never equal -2.

9. Originally Posted by helpless13
thanks i really appreciate your help
Be sure to see that the method that icemanfan showed you still works; you just have to be careful about your signs:

$\left\lvert2x + 1\right\rvert = x - 1$

$\Rightarrow\left(x\ge-\frac12\text{ and }2x + 1 = x - 1\right)\text{ or }\left(x<-\frac12\text{ and }-2x - 1 = x - 1\right)$

Considering the two cases separately:

$2x + 1 = x - 1\Rightarrow x = -2$, but this contradicts our requirement that $x\ge-\frac12,$ so it is not a solution.

$-2x - 1 = x - 1\Rightarrow x = 0$, but this contradicts our requirement that $x<-\frac12,$ so it is also not a solution.

These are the only two possibilities (from the trichotomy law), so the original equation has no solutions.

It is, however, always a good idea to check your values in the original equation, because unless you are very careful in performing operations you can easily end up with extraneous (and sometimes missing) solutions.

10. Originally Posted by angel.white
$|2x+1| = x-1$

You can pull the one out since it is always positive:

$|2x|+1 = x-1$

$|2x|-x = -2$

From here it should be clear that, as you have already discovered, there are no solutions, because no matter the value of x, |2x| will always be greater than it, and so |2x|-x will always be greater than or equal to zero, and thus can never equal -2.

You can only say that $|2x+1|\le |2x|+1$

Counter example :
$x=-\frac 13$

$|2x+1|=\frac 13$

$|2x|+1=\frac 53$ !

11. Originally Posted by Moo

You can only say that $|2x+1|\le |2x|+1$
$\left\lvert a+b\right\rvert\ne\left\lvert a\right\rvert + \left\lvert b\right\rvert,$
$\left\lvert a+b\right\rvert\le\left\lvert a\right\rvert + \left\lvert b\right\rvert$