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Linear Equation in 2 Variables

Simplify:3k-4...

Question

Simplify:
$(3 k - 4 r - 2 m)^{2} + (3 k + 4 r - 2 m)^{2}$

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Solution

Given expression $(3 k - 4 r - 2 m)^{2} + (3 k + 4 r - 2 m)^{2}$

Use the formula $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
By substitute $a = 3 k, b = - 4 r, c = - 2 m$ , we get
Thus, $(3 k - 4 r - 2 m)^{2} = (3 k)^{2} + (- 4 r)^{2} + (- 2 m)^{2} + 2 (3 k) (- 4 r) + 2 (- 4 r) (- 2 m) + 2 (- 2 m) (3 k)$
$\Rightarrow (3 k - 4 r - 2 m)^{2} = 9 k^{2} + 16 r^{2} + 4 m^{2} - 24 r k + 16 r m - 12 k m$ .......(1)
Similar way, $(3 k + 4 r - 2 m)^{2} = 9 k^{2} + 16 r^{2} + 4 m^{2} + 24 r k - 16 r m - 12 k m$ ......(2)
Adding equation(1) and (2), we get the required expression as
$(3 k - 4 r - 2 m)^{2} + (3 k + 4 r - 2 m)^{2} = 9 k^{2} + 16 r^{2} + 4 m^{2} - 24 r k + 16 r m - 12 k m + 9 k^{2} + 16 r^{2} + 4 m^{2} + 24 r k - 16 r m - 12 k m$
$= 18 k^{2} + 32 r^{2} + 8 m^{2} - 24 m k$

$∴ (3 k - 4 r - 2 m)^{2} + (3 k + 4 r - 2 m)^{2} = 18 k^{2} + 32 r^{2} + 8 m^{2} - 24 m k$

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