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CAT Previous Year Paper | CAT Linear and Quadratic Equations Questions | Question 16

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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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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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