If $3 a - 4 b = 7$ and $a b = 10$ find $9 a^{2} + 16 b^{2}$

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Solution

$3 a - 4 b = 7$ and $a b = 10$

$(3 a + 4 b)^{2} = (3 a - 4 b)^{2} + 4 (3 a) (4 b)$

$9 a^{2} + 16 b^{2} + 2 (3 a) (4 b) = (3 a - 4 b)^{2} + 4 (3 a) (4 b)$

$9 a^{2} + 16 b^{2} = (3 a - 4 b)^{2} + 24 a b$

Substitute given values

$9 a^{2} + 16 b^{2} = (7)^{2} + 24 (10)$

$= 49 + 240$

$= 289$ .

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