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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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