If x = 8 + √5 and y= 8-√5 then the value of x² + y² is:
138
We are given \(x = 8 + \sqrt{5}\) and \(y = 8 - \sqrt{5}\), and we need to find \(x^2 + y^2\).
Using the algebraic identity \((a+b)^2 + (a-b)^2 = 2(a^2 + b^2)\), with \(a = 8\) and \(b = \sqrt{5}\):
\(x^2 + y^2 = (8+\sqrt{5})^2 + (8-\sqrt{5})^2 = 2\left(8^2 + (\sqrt{5})^2\right)\)
\(= 2(64 + 5) = 2 \times 69 = 138\)
Verification by expansion:
\((8+\sqrt{5})^2 = 64 + 16\sqrt{5} + 5 = 69 + 16\sqrt{5}\)
\((8-\sqrt{5})^2 = 64 - 16\sqrt{5} + 5 = 69 - 16\sqrt{5}\)
Adding, the \(16\sqrt{5}\) terms cancel: \(69 + 69 = 138\).
Hence, the value of \(x^2 + y^2\) is 138.
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