Observation more than calculation
b = 3 + a 2 + a 2 + 3 3 − a 2 − a 2 − 3
The equation above holds true for real values a and b . Find b .
The answer is 0.
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3 solutions
I do not understand.
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If a 2 < 3 , then a 2 − 3 < 0 and a 2 − 3 is not real. If a 2 > 3 , then 3 − a 2 < 0 and 3 − a 2 is not real. Since b is real only when both 3 − a 2 and a 2 − 3 are real, which only occurs when a 2 = 3 . Then
b = 3 + a 2 + a 2 + 3 3 − a 2 − a 2 − 3 = 6 + 6 0 + 0 = 0
Simply, substitute 'a' for any number consistent throughout the equation. You'll get b = 0.
Since we are told that a and b are real, we are supposed to find b(real number) so we assume a value real value of a that'll give us a value of b that's a real number. So the only real value of a that give you a real b is √3, so b = 0.
For both 3 − a 2 and a 2 − 3 to be real, a 2 = 3 . Therefore b = 0 .