If $(a + b) = 5$ and $a b = 6$ ,find the value of $a^{2} + b^{2}$ ; $a > b$

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Solution

Given: $(a + b) = 5$ and $a b = 6$ ; where $a > b$
Since $(a + b)^{2} = a^{2} + b^{2} + 2 a b$
Substitute given values, we get
$(5)^{2} = a^{2} + b^{2} + 2 (6)$
$\Rightarrow 25 = a^{2} + b^{2} + 12$
$\Rightarrow a^{2} + b^{2} = 25 - 12$
$∴ a^{2} + b^{2} = 13$

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