# x=(sqrt(y^2)+1)/z

### x=(sqrt(y^2)+1)/z

x=(sqrt(y^2)+1)/z
we asume that x,y and z is > 0
what is y??
Guest

y =( zx -1)
Guest

### Re: x=(sqrt(y^2)+1)/z

i thought so too, but that was inncorrect Guest

### Re: x=(sqrt(y^2)+1)/z

Then equation must have some other meaning rather than simple algebra....
The sqrt (Y^2 ) is simply Y
Then Eqn as given is same as X = (Y + 1)/Z

Then ZX = Y+1

and Y = ZX -1
Guest

### Re: x=(sqrt(y^2)+1)/z

X=(√y^2+1)/z
Xz=√y^2+1
X^2z^2=y^2+1
X^2z^2-1=y^2
Y=√(x^2z^2-1)

TutorSOS

Posts: 5
Joined: Tue Sep 16, 2014 12:08 pm
Reputation: 1

### Re: x=(sqrt(y^2)+1)/z

tutorSOS thank you!!!! Can you help me with another one ?

5*x+z=(sqrt(y^2)+5)/z

and also need to get y
Guest

### Re: x=(sqrt(y^2)+1)/z

Hi TutorSOS .....I have to say I don't aggree with your working......

X = (sqrt(Y^2) + 1) / Z

You cross multiply by Z

XZ = (sqrt(Y^2) + 1) .....OK t here....

You then square both sides....I think...

X^2Z^2 = (Y^2) + 1 .....dont't aggree.....

How can this be??...Looks like you have forgotten about the +1....

The above line should be....

X^2Z^2 = (sqrt(Y^2) + 1)^2

this is same as..... X^2Z^2 = (sqrt(Y^2) + 1)(sqrt(Y^2) + 1)
this gives a totally different working out...

X = (sqrt(Y^2) + 1) / Z

cross multiply by Z

XZ = (sqrt(Y^2) + 1) .....OK

Bring over the +1
XZ - 1 = sqrt(Y^2)

Square both sides.....

(XZ - 1)^2 = Y^2

Now take sqrt of both sides.....

XZ - 1 = Y

But you don't really have to do all this squaring and square roots..

Notice that Sqrt(Y^2) is simply equal to Y and answer as done in earlier post...

I appologise if I have mistook some of your working....and would like to know if I am wrong can you explain where..??
Guest

### Re: x=(sqrt(y^2)+1)/z

Can you help me with another one ?

5*x+z=(sqrt(y^2)+5)/z

and also need to get y

Does this question read like this....??????

(5X)+Z =(sqrt(Y^2) + 5)/Z

Then Z(5X+Z) = Y + 5

So..... Z(5X+Z)-5 = Y
Guest

### Re: x=(sqrt(y^2)+1)/z

thank you guest but i tried that one and it was wrong Guest

### Re: x=(sqrt(y^2)+1)/z

If you "know it is wrong" then you must know the answer. And I assume the answer must be from a reputable source.
If that is the case then Post the answer you have and I will be able to see where I have gone wrong.

OR else I am not understanding the question or don't understand how you have written it on this text based forum.

To me they are simple algebra questions involving indices and roots.
Guest

### Re: x=(sqrt(y^2)+1)/z

Hi,
In the above problem I am taking
√(y^2+1) i.e. whole root of y^2+1
That’s why I am getting the answer in the above way.

TutorSOS

Posts: 5
Joined: Tue Sep 16, 2014 12:08 pm
Reputation: 1

### Re: x=(sqrt(y^2)+1)/z

5*x+z=√(y^2+5)/z
Some confusion is there in your problem, I am taking whole root of (y^2+5).
z(5*x+z)=√(y^2+5)
5*x*z+z^2=√(y^2+5)
Squaring on both sides
(5xz+z^2)^2=y^2+5
25x^2z^2+z^4+10xz^3=y^2+5
y^2=25x^2z^2+z^4+10xz^3-5
Now,
y=√(25x^2z^2+z^4+10xz^3-5)

TutorSOS

Posts: 5
Joined: Tue Sep 16, 2014 12:08 pm
Reputation: 1

### Re: x=(sqrt(y^2)+1)/z

The question as presented by Guest1 is as below.....
x=(sqrt(y^2)+1)/z
we asume that x,y and z is > 0
what is y??

It clearly shows the (y^2) enclosed by brackets and the sqrt before the bracket...thus the sqrt is for what is in the bracket only (Y^2).

The + 1 is outside the bracket for the sqrt but is within the bracket that enclosed the whole RHS of the equation numerator. So the whole RHS outer bracket is divided by Z

This is how Guest2 interpreted the question and answered it correctly giving Y = ZX -1

TutorSOS mis-interpreted or took down the question wrongly and therefore got the wrong answer

The second question presented by guest1 was as below....

5*x+z=(sqrt(y^2)+5)/z

and also need to get y

Clearly the brackets are in the same place so this is the same type of question involving sqrt(Y^2) only and Guest2 answered it correctly .... Y = Z(5X+Z)-5

And again TutorSOS took it down wrongly and got the wrong answer.
Guest

### Re: x=(sqrt(y^2)+1)/z

Very interesting for me! Well done!

palashoni

Posts: 2
Joined: Wed Oct 08, 2014 4:38 am
Reputation: 0