If \(x=a+b+\frac{(a-b)^2}{4 a+4 b}\) and \(y=\frac{a+b}{4}+\frac{a b}{a+b}\) then what is the value of (x - a)2 - (y - b)2 ?
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The question asks us to find the value of the expression \((x - a)^2 - (y - b)^2\), given the definitions of \(x\) and \(y\) in terms of \(a\) and \(b\). This problem involves simplifying complex algebraic expressions and potentially using algebraic identities like the difference of squares.
The given expressions are:
We need to calculate \((x - a)\) and \((y - b)\) first, then square them, and finally find the difference.
Let's first simplify the expression for \(x\). We can rewrite the denominator \(4a+4b\) as \(4(a+b)\).
\(x = a+b+\frac{(a-b)^2}{4(a+b)}\)
Consider the term \(\frac{(a-b)^2}{4(a+b)}\). We can expand the numerator and see if it relates to \(a+b\):
\(\frac{(a-b)^2}{4(a+b)} = \frac{a^2 - 2ab + b^2}{4(a+b)}\)
We can try rewriting the numerator in terms of \((a+b)^2\):
\(a^2 - 2ab + b^2 = (a^2 + 2ab + b^2) - 4ab = (a+b)^2 - 4ab\)
So, \(\frac{(a-b)^2}{4(a+b)} = \frac{(a+b)^2 - 4ab}{4(a+b)} = \frac{(a+b)^2}{4(a+b)} - \frac{4ab}{4(a+b)} = \frac{a+b}{4} - \frac{ab}{a+b}\).
Now substitute this back into the expression for \(x\):
\(x = a+b + \left(\frac{a+b}{4} - \frac{ab}{a+b}\right)\)
Using the simplified expression for \(x\):
\(x - a = \left(a+b + \frac{a+b}{4} - \frac{ab}{a+b}\right) - a\)
\(x - a = b + \frac{a+b}{4} - \frac{ab}{a+b}\)
Now let's calculate \((y - b)\) using the given expression for \(y\):
\(y = \frac{a+b}{4}+\frac{a b}{a+b}\)
\(y - b = \left(\frac{a+b}{4}+\frac{a b}{a+b}\right) - b\)
To combine the terms, we can find a common denominator for the last two terms:
\(\frac{ab}{a+b} - b = \frac{ab - b(a+b)}{a+b} = \frac{ab - ab - b^2}{a+b} = \frac{-b^2}{a+b}\)
So, the expression for \((y - b)\) becomes:
\(y - b = \frac{a+b}{4} - \frac{b^2}{a+b}\)
We need to calculate \((x - a)^2 - (y - b)^2\). This is a difference of squares, which can be factored as \(((x - a) - (y - b))((x - a) + (y - b))\).
Substitute the expressions for \((x-a)\) and \((y-b)\):
\((x - a) - (y - b) = \left(b + \frac{a+b}{4} - \frac{ab}{a+b}\right) - \left(\frac{a+b}{4} - \frac{b^2}{a+b}\right)\)
Removing the parentheses:
\((x - a) - (y - b) = b + \frac{a+b}{4} - \frac{ab}{a+b} - \frac{a+b}{4} + \frac{b^2}{a+b}\)
Cancel the terms \(\frac{a+b}{4}\) and \(-\frac{a+b}{4}\):
\((x - a) - (y - b) = b - \frac{ab}{a+b} + \frac{b^2}{a+b}\)
Combine the last two terms with a common denominator \((a+b)\):
\((x - a) - (y - b) = b + \frac{b^2 - ab}{a+b}\)
Combine \(b\) with the fraction:
\((x - a) - (y - b) = \frac{b(a+b) + b^2 - ab}{a+b} = \frac{ab + b^2 + b^2 - ab}{a+b} = \frac{2b^2}{a+b}\)
Substitute the expressions for \((x-a)\) and \((y-b)\):
\((x - a) + (y - b) = \left(b + \frac{a+b}{4} - \frac{ab}{a+b}\right) + \left(\frac{a+b}{4} - \frac{b^2}{a+b}\right)\)
Removing the parentheses:
\((x - a) + (y - b) = b + \frac{a+b}{4} - \frac{ab}{a+b} + \frac{a+b}{4} - \frac{b^2}{a+b}\)
Group like terms:
\((x - a) + (y - b) = b + 2\left(\frac{a+b}{4}\right) - \left(\frac{ab}{a+b} + \frac{b^2}{a+b}\right)\)
\((x - a) + (y - b) = b + \frac{a+b}{2} - \frac{ab+b^2}{a+b}\)
Factor \(b\) from the numerator of the last fraction:
\((x - a) + (y - b) = b + \frac{a+b}{2} - \frac{b(a+b)}{a+b}\)
Assuming \(a+b \neq 0\), we can cancel \((a+b)\) in the last term:
\((x - a) + (y - b) = b + \frac{a+b}{2} - b\)
\((x - a) + (y - b) = \frac{a+b}{2}\)
Using the difference of squares formula:
\((x - a)^2 - (y - b)^2 = ((x - a) - (y - b))((x - a) + (y - b))\)
Substitute the results from the previous steps:
\((x - a)^2 - (y - b)^2 = \left(\frac{2b^2}{a+b}\right) \left(\frac{a+b}{2}\right)\)
Assuming \(a+b \neq 0\), we can multiply the fractions and cancel terms:
\((x - a)^2 - (y - b)^2 = \frac{2b^2 \cdot (a+b)}{(a+b) \cdot 2}\)
\((x - a)^2 - (y - b)^2 = b^2\)
The value of \((x - a)^2 - (y - b)^2\) is \(b^2\).
| Step | Expression / Calculation | Result |
|---|---|---|
| Simplify fraction in \(x\) | \(\frac{(a-b)^2}{4(a+b)}\) | \(\frac{a+b}{4} - \frac{ab}{a+b}\) |
| Calculate \(x - a\) | \(b + \left(\frac{a+b}{4} - \frac{ab}{a+b}\right)\) | \(b + \frac{a+b}{4} - \frac{ab}{a+b}\) |
| Calculate \(y - b\) | \(\frac{a+b}{4} + \frac{ab}{a+b} - b\) | \(\frac{a+b}{4} - \frac{b^2}{a+b}\) |
| Calculate \((x-a) - (y-b)\) | \(\left(b + \frac{a+b}{4} - \frac{ab}{a+b}\right) - \left(\frac{a+b}{4} - \frac{b^2}{a+b}\right)\) | \(\frac{2b^2}{a+b}\) |
| Calculate \((x-a) + (y-b)\) | \(\left(b + \frac{a+b}{4} - \frac{ab}{a+b}\right) + \left(\frac{a+b}{4} - \frac{b^2}{a+b}\right)\) | \(\frac{a+b}{2}\) |
| Final Result | \(((x-a) - (y-b))((x-a) + (y-b))\) | \(b^2\) |
This problem effectively used the difference of squares identity: \(A^2 - B^2 = (A - B)(A + B)\). By identifying \((x-a)\) as \(A\) and \((y-b)\) as \(B\), we were able to simplify the final calculation significantly. Problems involving complex expressions can often be solved by simplifying parts of the expression first and looking for common terms or structures. The key simplification \(\frac{(a-b)^2}{4(a+b)} = \frac{a+b}{4} - \frac{ab}{a+b}\) was crucial in expressing \((x-a)\) and \((y-b)\) in forms that combined nicely in the final step.
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