Question

x^2+y^2 = 42 , xy=8 find (3x-3y)^2​

Answer 1

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Given:

• $x^2 + y^2 = 42$
• $xy = 8$

To Find:

• $Value \:of\: (3x - 3y)^2$

Solution:

$\implies (3x - 3y)^2 \\ \implies 3^2*(x - y)^2 \\ \implies 9(x^2 + y^2 - 2xy) \\ \implies 9(42 - 2(8)) \\ \implies 9(42 - 16) \\ \implies 9\times (26) \\ \bold {\implies 234}$

Formula Used:

• $(a - b)^2 = a^2 + b^2 - 2ab$

Hope it helps!!!