Question

If x and y are two positive numbers such that 4x2+y2=40 and xy=6,find the value of 2x+y

Answer 1

$(2x + y) { }^{2} = 4 {x}^{2} + {y}^{2} + 4xy \\ (2x + y) {}^{2} = 40 + 4(6) \\ (2x + y) = + - 8 \\ sincex \: and \: y \: are \: positive \: integers \: \\ 2x + y = 8$

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Answer 2

$4x {}^{2} + y {}^{2} = 40 \\ xy = 6 \\ (2x + y) {}^{2} = (2x) {}^{2} + (y) {}^{2} + 2 \times 2x \times y \\ = 4x {}^{2} + y {}^{2} + 4xy \\ = 40 + 4 \times 6 \\ = 40 + 24 \\ = 64$
Now,
$(2x + y) {}^{2} = 64 \\ 2x + y = \sqrt{64} \\ 2x + y = 8$