If $α$ and $β$ are the zeroes of the polynomial $f (x) = x^{2} - 6 x + k$ , find the value of $k$ , such that $α^{2} + β^{2} = 40$

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Solution

Given: $α, β$ are the zeroes of the polynomial $f (x) = x^{2} - 6 x + k$ and $α^{2} + β^{2} = 40$
To find the value of $k$
Sol: We know,
$α + β = 6, α β = k$
And given $α^{2} + β^{2} = 40$
$⟹ (α + β)^{2} - 2 α β = 40$ as $(α + β)^{2} = α^{2} + β^{2} + 2 α β$
$⟹ 6^{2} - 2 (k) = 40 ⟹ 36 - 2 k = 40 ⟹ 2 k = - 4 ⟹ k = - 2$

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