Prove that A.M. of the roots of $x^{2} - 2 a x + b^{2} = 0$ is equal to the geometric mean of the roots of the equation $x^{2} - 2 b x + a^{2} = 0$ , and vice-versa.

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Solution

$α + β = 2 a o r \frac{α + β}{2} = a,$
$γ δ = a^{2}$
$∴ \sqrt{γ δ} = a$
$∴$ A.M. of roots of 1st = G.M. of roots of 2nd.

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