If $a + b = 10, a b = 32$ Find $a^{2} + b^{2}, a - b$ & $a^{2} - b^{2}$

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Solution

Given $a + b = 10 - - - (1)$
$a b = 32 - - - (2)$
$(a + b)^{2} = a^{2} + b^{2} + 2 a b$
$(10)^{2} = a^{2} + b^{2} + 64$
$a^{2} + b^{2} = 100 - 84 = 36$
$∴ a^{2} + b^{2} = 36 - - - (3)$

Now $(a - b)^{2} = a^{2} + b^{2} - 2 a b$
$a - b^{2} = 36 - 64$
$(a - b)^{2} = - 28$
$a - b = \sqrt{28} = + 2 \sqrt{7} i - - - - (4)$

Now $a^{2} - b^{2} = (a + b) (a - b) = 10 (2 \sqrt{7} i)$
$a^{2} - b^{2} = 20 \sqrt{7} i - - - - - (5)$

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