# Thread: Help again with Modulus & Irrational Equations

1. ## Help again with Modulus & Irrational Equations

Can anyone help with this as I am stumped:

1 + square root of x = square root of 3(x-1)

The answer is x=4 but I can't get it.
Also can you tell me how to do the square root symbol etc when posting a thread.

Thanks and Kind Regards

2. Hello,

Square the whole thing...

Then simplify what you can and subsitute $u=\sqrt{x}$

For the square root : between [math ] and [/tex] do \sqrt
See the code by clicking on my square root

3. Thanks for your help moo, unfortunately I am totally lost with regard to the equation but thank you for helping me with the symbols!!!!

Regards

4. Originally Posted by larsson
Can anyone help with this as I am stumped:

1 + square root of x = square root of 3(x-1)

The answer is x=4 but I can't get it.
Also can you tell me how to do the square root symbol etc when posting a thread.

Thanks and Kind Regards
as Moo directed.

$1 + \sqrt{x} = \sqrt{3(x - 1)}$ ..........square both sides

$\Rightarrow (1 + \sqrt{x})^2 = (\sqrt{3(x - 1)})^2$ .........simplify

$\Rightarrow 1 + 2 \sqrt{x} + x = 3(x - 1)$

can you continue?\

5. Ok, i'll go through it

$1+\sqrt{x}=\sqrt{3(x-1)}$

If we square both sides :

$\underbrace{(1+\sqrt{x})^2}_{1+x+2\sqrt{x}}=3(x-1)=3x-3$

$1+x+2 \sqrt{x}=3x-3$

$2 \sqrt{x}=2x-4$

$x-\sqrt{x}-2=0$

Substitute $u=\sqrt{x}$

$x=(\sqrt{x})^2=u^2$

So it's the same as solving for u (positive) in :

$u^2-u-2=0$

This is $(u-2)(u+1)=0$

So u is either equal to 2 or to -1.

But $u=\sqrt{x}>0$

So u=2.

Thus x=...

6. If we square both sides :

Its up to here that I could get to and understand. The rest of it has me totally lost, I am still not sure how the answer is 4.

I really appreciate all the help you have all given me and I hope I am not annoying you all

Regards

7. Originally Posted by larsson

If we square both sides :

Its up to here that I could get to and understand. The rest of it has me totally lost, I am still not sure how the answer is 4.

I really appreciate all the help you have all given me and I hope I am not annoying you all

Regards