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Question 16 : If u² + (u−2v−1)² = −4v(u + v), then what is the value of u + 3v?

1. \\frac{1}{4})
2. \\frac{1}{2})
3. 0
4. -\\frac{1}{4})

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Video Explanation

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Explanatory Answer

Method of solving this CAT Question from Linear & Quadratic Equations

Given expression, u² + (u − 2v − 1)² = −4v(u + v)
On expanding using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, we get
u² + u² + 4v² + 1 - 4uv - 2u + 4v = - 4vu - 4v²
2u² + 4v² - 2u + 4v + 4v² + 1 = 0
On rearranging and taking common we get,
2(u² - u) + 2 (4v² + 2v) = - 1
Add values on both sides to complete the squares inside parenthesis
2(u² – u + \\frac{1}{4})) + 2(4v² + 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 u² + (u−2v−1)² = −4v(u + v), then what is the value of u + 3v?"

Hence, the answer is -\\frac{1}{4}).

Choice D is the correct answer.

 

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