eq b $. Find the value of $ rac{a + b}{a - b} + rac{a - b}{a + b} $.
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- Equating coefficients: $ 2c = 2b y + 4c $ for all $ y $, which implies $ b = 0 $ and $ c = 0 $. Thus, $ f(x) = ax^2 $. Verification confirms this satisfies the equation. Hence, all solutions are quadratic functions of the form $ f(x) = kx^2 $.
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- Question: Let $ a $ and $ b $ be complex numbers such that $ |a| = |b| = 1 $ and $ a
- Solution: Let $ S = rac{a + b}{a - b} + rac{a - b}{a + b} $. Combine the fractions:
- S = rac{(a + b)^2 + (a - b)^2}{(a - b)(a + b)} = rac{2a^2 + 2b^2}{a^2 - b^2}.
- Since $ |a| = |b| = 1 $, $ a \overline{a} = 1 $ and $ b \overline{b} = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overline{S} $:
