Pythagorean theorem

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rachelmaddie

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Can someone please check my work?
Applying the Pythagoras theorem c^2 = a^2 + b^2
c = (10x + 15y)
a = (6x + 9y)
b = (8x + 12y)

a
(10x + 15y)^2 = (6x + 9y)^2 + (8x + 12y)^2

Second, transform each side of the equation to determine if it is an identity.

b
100x^2 + 150xy + 225y^2 = 36x^2 + 54xy + 81y^2 + 64x^2
100x^2 + 150xy + 225^2 = 100x^2 + 150xy + 225^2

The left side is equal to the right side, therefore they are an identity.

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Not quite, please double check the red text below...

(10x + 15y)^2 = 100x^2 + 150xy + 225y^2 ??

(Similar mistakes on the RHS too)
 
(a+b)^2 = (a+b)(a+b) = a^2 + a*b + b*a + b^2 = a^2 + 2*a*b + b^2
 
Strange. "Write an equation"? The only equation I can see here is 2x+3y not = 0.
9+16=25 is a simple identity.
I don't know what you are supposed to learn in this problem.
 
I was always told to deal with the LHS and RHS separately, so I'd start with

Consider the LHS
(10x + 15y)^2
100x^2 + 2*150xy + 225y^2
100x^2 + 300xy + 225y^2

Consider the RHS
(6x + 9y)^2 + (8x + 12y)^2
...
(I'll leave this work for you!)
...
=100x^2 + 300xy + 225y^2

And then finish off with the conclusion statement "LHS = RHS therefore the identity is true"
 
(a+b)^2 = a^2 + b^2 +2ab

So
(6x + 9y)^2 = 36x^2 + 82y^2 + 108xy
and (8x + 12y)^2 = 64x^2 + 144y^2 + 192xy
 
(6x + 9y)^2 = 36x^2 + 81y^2 + 108xy
(8x + 12y)^2 = 64x^2 + 144y^2 + 192xy

Now add them. Get 100x^2 + 225y^2 + 300xy. Note this equals (10x+15y)^2.

So yes, (6x + 9y)^2 + (8x + 12y)^2 = (10x+15y)^2.
 
\(\displaystyle (6x + 9y)^2 \ + \ (8x + 12y)^2 \ \ vs. \ (10x + 15y)^2\)

\(\displaystyle [3(2x + 3y)]^2 \ + \ [4(2x + 3y)]^2 \ \ vs. \ [5(2x + 3y)]^2\)

\(\displaystyle (3^2 + 4^2)(2x + 3y)^2 \ \ vs. \ 5^2(2x + 3y)^2\)

\(\displaystyle (9 + 16)(2x + 3y)^2 \ \ vs. \ 25(2x + 3y)^2\)

\(\displaystyle 25(2x + 3y)^2 \ = \ 25(2x + 3y)^2\)
 
lookagain said:
\(\displaystyle (6x + 9y)^2 \ + \ (8x + 12y)^2 \ \ vs. \ (10x + 15y)^2\)

\(\displaystyle [3(2x + 3y)]^2 \ + \ [4(2x + 3y)]^2 \ \ vs. \ [5(2x + 3y)]^2\)

\(\displaystyle (3^2 + 4^2)(2x + 3y)^2 \ \ vs. \ 5^2(2x + 3y)^2\)

\(\displaystyle (9 + 16)(2x + 3y)^2 \ \ vs. \ 25(2x + 3y)^2\)

\(\displaystyle 25(2x + 3y)^2 \ = \ 25(2x + 3y)^2\)
Click to expand...
Is this another way that the second step can be written?
 
What you have is already an identity. I mentioned
9+16=25
32 + 42 = 52
Now multiply through by (2x+3y)2

I think this problem is just not well-written. Is it from a book?
 
Singleton said:
What you have is already an identity. I mentioned
9+16=25
32 + 42 = 52
Now multiply through by (2x+3y)2

I think this problem is just not well-written. Is it from a book?
Click to expand...
No, this is part of my lesson from my teacher. I’m still not sure if the work I’ve shown above is correct or not?
 
Hmm... ok let's see if we can figure out what you teacher wants. "Transform each side"... ok if you want to multiply it out. RHS: You did it right for the (6x+9y)2 term. Then it looks like for the (8x+12y)2 term it got cut off. (Intermediate step). Jomo and Cubist above wrote out the missing part for you. It's fine at the end.
 
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