If the sum of the square of the zeroes of the quadratic polynomial $x^{2} - 8 x + k$ is $40,$ find the value of K.

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Solution

Let the roots of $x^{2} - 8 x + k = 0$ be $a, b$
$⟹ a + b = 8, a b = k$
Given $a^{2} + b^{2} = 40 ⟹ (a + b)^{2} - 2 a b = 40$
$⟹ 64 - 2 k = 40 ⟹ 2 k = 24 ⟹ k = 12$

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