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For all real numbers a and b, 2a•b=a^2 + b^2

User Klaasman
by
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2 Answers

2 votes
hmm

one solutionm is if both a and b were 0
the resulting equation would be 0=0 which is true
(0,0) is one solution

another is if both a and b are 1
resulting equation would be 2=2 which is true

I just graphed it and the solution is all values such that a=b
so
(a,b)
(1,1)
(2,2)
(π,π)
etc
(1,1) is a solution
User Timbo White
by
8.3k points
0 votes
That's not true at all.

If a=2 and b=3 . . . then

2 · a · b = 12
but
a² + b² = 13 .

This doesn't satisfy your statement, because 12 and 13 are not equal.

Right off hand, it looks to me as if your statement is true
ONLY if
|a| = |b| ('a' = plus or minus 'b')

and NOT for any other pairs of numbers.
User Colin Alworth
by
7.9k points

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