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Question 16 : If u^{2} + (u−2v−1)^{2} = −4v(u + v), then what is the value of u + 3v?

- \\frac{1}{4})
- \\frac{1}{2})
- 0
- -\\frac{1}{4})

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Given expression, u^{2} + (u − 2v − 1)^{2} = −4v(u + v)

On expanding using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca, we get

u^{2} + u^{2} + 4v^{2} + 1 - 4uv - 2u + 4v = - 4vu - 4v^{2}

2u^{2} + 4v^{2} - 2u + 4v + 4v^{2} + 1 = 0

On rearranging and taking common we get,

2(u^{2} - u) + 2 (4v^{2} + 2v) = - 1

Add values on both sides to complete the squares inside parenthesis

2(u^{2} – u + \\frac{1}{4})) + 2(4v^{2} + 2v + \\frac{1}{4})) = -1 + \\frac{1}{2}) + \\frac{1}{2})

2(u - \\frac{1}{2}))^{2} + 2(2v + \\frac{1}{2}))^{2} = 0

Since they both are squares, they can’t be negative

So, u - \\frac{1}{2}) = 0 and 2v + \\frac{1}{2}) = 0

u = \\frac{1}{2}) and v = −\\frac{1}{4})

U+3v = \\frac{1}{2}) - \\frac{3}{4}) = -\\frac{1}{4})

The question is **"If u ^{2} + (u−2v−1)^{2} = −4v(u + v), then what is the value of u + 3v?" **

Choice D is the correct answer.

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