Pythagorean theorem

Singleton said:
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.
Click to expand...
Where is the missing part?
 
\((a+b)^2=a^2+2ab+b^2\)
\((6x+4y)^2=(6x)^2+2(6x)(4y)+(4y)^2=36x^2+48xy+16y^2\)
NOW
\((3t-5s)^2=(3t)^2+2(3t)(-5s)+(-5s)^2=9t^2-30ts+25s^2\)

Just calm down do these one term at a time.
 
Singleton said:
You had 81y^2 then in the next line became 225y^2.
Click to expand...
36x^2 + 108xy + 81y^2 + 64x^2 + 192 xy + 144 y = 100x^2 + 300xy + 225y^2
36x^2 + 64x^2 + 108xy + 192xy + 81y^2 + 144y^2 = 100x^2 + 300xy + 225y^2
100x^2 + 300xy + 225y^2 = 100x^2 + 300xy + 225y^2

How is this?
 
rachelmaddie said:
Is this another way that the second step can be written?
Click to expand...


I had not gotten back with you. No, it is not another way the second step
can be written. It is its own method. It is a slick way of doing the problem
that avoids expanding a binomial raised to an exponent. The problem
shows that expanding is not a requirement, depending on the route taken.

However, if you choose, ignore my posts and follow the more recent ones
for you.
 
lookagain said:
I had not gotten back with you. No, it is not another way the second step
can be written. It is its own method. It is a slick way of doing the problem
that avoids expanding a binomial raised to an exponent. The problem
shows that expanding is not a requirement, depending on the route taken.

However, if you choose, ignore my posts and follow the more recent ones
for you.
Click to expand...
Can you please check this for me?
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
(6x + 9y)^2 + (8x + 12y)^2 + vs. (10x + 15y)^2
[3(2x + 3y)]^2 + [4(2x + 3y)]^2 + vs. [5(2x + 3y)]^2
(3^2 + 4^2)(2x + 3y)^2 vs. 5^2(2x + 3y)^2
(9 + 16)(2x + 3y^2) vs. 25(2x + 3y)^2
25(2x + 3y)^2 = 25(2x + 3y)^2

The left side is equal to the right side, therefore they are an identity.
 
rachelmaddie said:
Can you please check this for me?

(6x + 9y)^2 + (8x + 12y)^2 + vs. (10x + 15y)^2 . . . . . No plus sign before "vs."
[3(2x + 3y)]^2 + [4(2x + 3y)]^2 + vs. [5(2x + 3y)]^2 . . . . . See the comment above.
(3^2 + 4^2)(2x + 3y)^2 vs. 5^2(2x + 3y)^2
(9 + 16)(2x + 3y^2) vs. 25(2x + 3y)^2 . . . . . The second close parenthesis needs
to be placed immediately to the right of the y-variable.
25(2x + 3y)^2 = 25(2x + 3y)^2
Click to expand...


The above appear to be typos.
 
lookagain said:
The above appear to be typos.
Click to expand...

b
(6x + 9y)^2 + (8x + 12y)^2 + vs. (10x + 15y)^2
[3(2x + 3y)]^2 + [4(2x + 3y)]^2 + vs. [5(2x + 3y)]^2
(3^2 + 4^2)(2x + 3y)^2 vs. 5^2(2x + 3y)^2
(9 + 16)(2x + 3y)^2 vs. 25(2x + 3y)^2
25(2x + 3y)^2 = 25(2x + 3y)^2

Hows this?
 
lookagain said:
The above appear to be typos.
Click to expand...
rachelmaddie said:
b
(6x + 9y)^2 + (8x + 12y)^2 + vs. (10x + 15y)^2
[3(2x + 3y)]^2 + [4(2x + 3y)]^2 + vs. [5(2x + 3y)]^2
(3^2 + 4^2)(2x + 3y)^2 vs. 5^2(2x + 3y)^2
(9 + 16)(2x + 3y)^2 vs. 25(2x + 3y)^2
25(2x + 3y)^2 = 25(2x + 3y)^2

Hows this?
Click to expand...


You also need to get rid of those extra plus signs in your top two lines
just before "vs." as I already stated in my prior post.
 
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