If $a - b = 6$ and $a b = 16$ , then find $a^{2} + b^{2}$ .

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Solution

The correct option is B 68
Formula ${(a - b)}^{2} = a^{2} + b^{2} - 2 a b$
$=> a^{2} + b^{2} = {(a - b)}^{2} + 2 a b$
Given, $a - b = 6$ and $a b = 16$
So, $a^{2} + b^{2} = {(a - b)}^{2} + 2 a b = 6^{2} + 2 \times 16 = 36 + 32 = 68$

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If a - b = 6 and ab = 16; find $a^{2} + b^{2}$

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