if a + bi = (x + i)² / 2x² + 1 , prove that a² + b² = (x² + 1 )² / (2x² + 1)²
Answers
Answer:
answer for the given problem is given
Answer:
The question implies the values for a and b will be integers; however, I don’t believe a AND be will be an integer - because the destination form requires an (x+a)² term.
3(x+a)² + b, expanded is:
3x² + 6ax + 3a² + b
And the original equation is 3x² - 12x - 11.
Setting them equal:
3x² + 6ax + 3a² + b = 3x² - 12x - 11
Less 3x² from both sides:
3x² + 6ax + 3a² + b - 3x² = 3x² - 12x - 11 - 3x²
6ax + 3a² + b = 12x - 11
Note the x?
That’s the problem - it means either a or b will be defined in terms of x.
Let’s say we solve for b …
6ax + 3a² + b = 12x - 11 (given)
6ax + 3a² = 12x - 11 - b (move b to the right)
6ax + 3a² - 12x + 11 = - b (move non-b to the left)
It’s even uglier solving for a …
6ax + 3a² + b = 12x - 11 (given)
6ax + 3a² = 12x - 11 - b (put all non-a terms on the right)
3 (a + 1)² - 3 = 12x - 11 - b (because we have an “a²” term we should probably complete the square)
3 (a + 1)² = 12x - 11 - b + 3 = 12x - b - 8 (move the three)
(a + 1)² = (12x - b - 8)/3 (divide by the three)
a + 1 = square root[ (12x - b - 8)/3 ] (square root)
a = square root[ (12x - b - 8)/3 ] - 1 (and less 1)
So you can DO it - but if the expectation is two integers - you’re only getting that if x ix 0.
In which case, a = 0, b = -11.............................................................========================================================================================================================================================================