# Quick question: x^2-y^2=16 (find x)

## Homework Statement

x^2-y^2=16
isolate the variable x

square roots

## The Attempt at a Solution

x^2=y^2+16

x=y+4?

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BvU
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Because your $x^2 = (y+4)^2 \ne y^2+16$

• Math_QED

if i squared both sides it would just be the individual answer

BvU
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What's the full problem statement ? 'Isolate' doesn't sound like a question with an answer ...

that was the problem..it just said isolate the variable x..

NO! This is a fundamental mistake. $\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$

What's the full problem statement ? 'Isolate' doesn't sound like a question with an answer ...
Isolate probably means solve for x in terms of y.

said something like it is understood to be positive or something like that

we were solving equations where you square both sides

BvU
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that was the problem..it just said isolate the variable x..
In that case you are done when you write $x^2 = y^2 + 16$ : it has x in isolation on the left hand side of the $=$ sign.

• Mastermind01
it cant be a square

Poster has been reminded (too late) not to post answers to schoolwork questions on the PF
it cant be a square
So the answer is just $x = \sqrt{y^2 + 16}$ NOT $y+4$

BvU
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Rapid fire of posts from three sides ...

Ok, semantics: isolate means write it so that it says $x =$

What do you have to keep in mind when going from $x^2 = ...$ to $x = ...$ ?

• Mastermind01
why doesnt that reduce?

BvU
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So the answer is just $x = \sqrt{y^2 + 16}$ NOT $y+4$
Watch out MM ! PF hates direct answers. it ruins the learning experience • berkeman and Mastermind01
learning experience means frustration and anger and deletion of account for me...

why doesnt that reduce?
Let's keep it cool. What's $(a+b)^2$?

a^2+b^2

BvU
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why doesnt that reduce?
It does reduce, but you have to be careful...
learning experience means frustration and anger and deletion of account for me...
I recognise the frustration and anger. Don't get carried away, sit back and think a bit now and then before firing of another post and the 'deletion of account' may be replaced by 'satisfaction from deeper udnerstanding' • Mastermind01
a^2+b^2
NO. Let's multiply it out. $(a+b)(a+b)$ Do you know the distributive law of multiplication?

• sobergeek23 and BvU
Newbie poster has been reminded (too late) not to give answers to homework problems here on the PF.
Let's take it step by step.
First things first, x2 + y2 is NOT equal to (x+y)2
That is because if we expand the second to (x+y)*(x+y) it's x*x + x*y + y*x + y*y which is equal to x2+2*x*y+y2.
This is the basic calculus formula for quadratic equations, which are equations to the power of 2.
Now, x2 + y2 is not the same as (x+y)2, using the rule above.
With another rule, we can expand x2 - y2 into this: (x-y)*(x+y), and to prove this we do the same as (x+y)*(x+y).
So: (x-y)*(x+y) is: x*x + x*y -y*x - y2 (plus times minus equals minus).
Therefore: x2 - y2.
So, going forward to your question,
x2 - y2 = 16 is as follows:
x2 = 16 + y2
x = √(16 + y2)

I hope I was able to shed some light for you.

• sobergeek23
BvU
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Hello G10, That has come by already and was chastized as giving a direct answer (and a wrong one to boot). Please be a bit more careful next time.

• berkeman
NO. Let's multiply it out. $(a+b)(a+b)$ Do you know the distributive law of multiplication?
yea i got it..is it because the plus sign is in there?

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is it aginst forum policy to give out answers?

• berkeman