# Thread: [SOLVED] Solving for A,B,C in context of Pythagorean Theorem

1. ## [SOLVED] Solving for A,B,C in context of Pythagorean Theorem

The equation is:
A^2 + B^2 = C^2

a. How would one solve for A?
b. How would one solve for B?
c. How would one solve for C?

--
My answers: (I'm almost certain I have these wrong)
a.
A^2 = C^2 - B^2
A = sqrt(C - B)
b.
B^2 = C^2 - A^2
B = sqrt(C -A)
c.
C^2 = A^2 + B^2
C = sqrt(A + B)

If I did, what properties did I misused? (I honestly can't tell)

2. Originally Posted by Masterthief1324
The equation is:
A^2 + B^2 = C^2

a. How would one solve for A?
b. How would one solve for B?
c. How would one solve for C?

--
My answers: (I'm almost certain I have these wrong)
a.
A^2 = C^2 - B^2
A = sqrt(C - B)
b.
B^2 = C^2 - A^2
B = sqrt(C -A)
c.
C^2 = A^2 + B^2
C = sqrt(A + B)

If I did, what properties did I misused? (I honestly can't tell)
on the right hand sides of your equations, where did the squares go when you took the square root?

i assume you are ignoring the negative square roots since we are working in the context of Pythagoras' theorem, and hence are assuming $\displaystyle A,B,C \ge 0$?

3. Originally Posted by Jhevon
on the right hand sides of your equations, where did the squares go when you took the square root?

i assume you are ignoring the negative square roots since we are working in the context of Pythagoras' theorem, and hence are assuming $\displaystyle A,B,C \ge 0$?
Yes.
A,B,C > 0

I figured that taking the sqrt of B^2 + C^2 would cancel out the power of 2.

Ohh wait, that's not a property is it? Why not? I figured that if Sqrt of X^2 = X and Y^2 = Y, then sqrt (x^2 + y^2) = x + y. I know for a fact that this isn't true - but why isn't it?

It should be ..
A = sqrt (C^2 - B^2)
B = sqrt (C^2 - A^2)
C = sqrt (A^2 + B^2)

4. Originally Posted by Masterthief1324
Yes.
A,B,C > 0

I figured that taking the sqrt of B^2 + C^2 would cancel out the power of 2.

Ohh wait, that's not a property is it? Why not? I figured that if Sqrt of X^2 = X and Y^2 = Y, then sqrt (x^2 + y^2) = x + y. I know for a fact that this isn't true - but why isn't it?

It should be ..
A = sqrt (C^2 - B^2)
B = sqrt (C^2 - A^2)
C = sqrt (A^2 + B^2)
Now you are correct. and yes $\displaystyle \sqrt{x^2 + y^2} \ne x + y$ in general, as you said. this is easy to see. assume $\displaystyle \sqrt{x^2 + y^2} = x + y$. now get rid of the square root on the left by squaring both sides, you get

$\displaystyle x^2 + y^2 = (x + y)^2 = x^2 + 2xy + y^2$ !

obviously this makes no sense, not unless $\displaystyle x$ and/or $\displaystyle y$ is zero. so our original assumption that the equation holds must have been an error.

5. Originally Posted by Jhevon
Now you are correct. and yes $\displaystyle \sqrt{x^2 + y^2} \ne x + y$ in general, as you said. this is easy to see. assume $\displaystyle \sqrt{x^2 + y^2} = x + y$. now get rid of the square root on the left by squaring both sides, you get

$\displaystyle x^2 + y^2 = (x + y)^2 = x^2 + 2xy + y^2$ !

obviously this makes no sense, not unless $\displaystyle x$ and/or $\displaystyle y$ is zero. so our original assumption that the equation holds must have been an error.
$\displaystyle x^2 + y^2 = (x + y)^2 = x^2 + 2xy + y^2$
Right so this doesn't make sense. It should be ..

$\displaystyle x^2 + y^2 \ne (x + y)^2 = x^2 + 2xy + y^2$

Which means you can't factor out common exponents of two sums by taking the sqrt of the sums of two different expression with exponents. However, is it possible to factor out sqrt only when you have a product of two expressions with either common or different exponents?

So ...
$\displaystyle C^2 = A^2 + B^2 \ne C = \sqrt{A} + \sqrt{B}$

but

$\displaystyle (C^2=A^2B^2) = (C=AB)$

6. Originally Posted by Masterthief1324
$\displaystyle x^2 + y^2 = (x + y)^2 = x^2 + 2xy + y^2$
Right so this doesn't make sense. It should be
$\displaystyle x^2 + y^2 \ne (x + y)^2 = x^2 + 2xy + y^2$ !
exactly.

the moral of the story: powers to not distribute across sums. (you can think of the square root as the 1/2 power). if you raise a sum to any power other than 1, the power does not apply to each term in the sum. that is, in general, $\displaystyle (a + b)^n \ne a^n + b^n$ for a sum of two terms.

7. Quote:
Originally Posted by Jhevon
Now you are correct. and yes in general, as you said. this is easy to see. assume . now get rid of the square root on the left by squaring both sides, you get

!

obviously this makes no sense, not unless and/or is zero. so our original assumption that the equation holds must have been an error.

Right so this doesn't make sense. It should be ..

Which means you can't factor out common exponents of two sums by taking the sqrt of the sums of two different expression with exponents. However, is it possible to factor out sqrt only when you have a product of two expressions with either common or different exponents?

So ...

but

8. Originally Posted by Masterthief1324
Quote:
Originally Posted by Jhevon
Now you are correct. and yes in general, as you said. this is easy to see. assume . now get rid of the square root on the left by squaring both sides, you get

!

obviously this makes no sense, not unless and/or is zero. so our original assumption that the equation holds must have been an error.

Right so this doesn't make sense. It should be ..

Which means you can't factor out common exponents of two sums by taking the sqrt of the sums of two different expression with exponents. However, is it possible to factor out sqrt only when you have a product of two expressions with either common or different exponents?

So ...

but

Provided all the variables are non-negative, yes. i think what you want to say is that you can distribute powers across products, but not sums. this is true.

(your first equation wouldn't make sense anyway )

i think you wanted to say: $\displaystyle C^2 = A^2 + B^2$ does not mean $\displaystyle C = \sqrt {A^2} + \sqrt {B^2} = A + B$, which is true. this is false.

but, provided A, B, C are non-negative, $\displaystyle C^2 = A^2 B^2 \implies C = AB$