# CAT 2018 Question Paper | Quants Slot 1

###### 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. $$frac{1}{4}$ 2. $\frac{1}{2}$ 3. 0 4. -$\frac{1}{4}$ ## 🎊 Independance Day Discount! Upto 4,000 off on CAT Online Courses Valid until 16th Aug #### 2IIM : Best Online CAT Coaching. ### Video Explanation > ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer ##### Method of solving this CAT Question from Linear & Quadratic Equations Given expression, u2 +$u − 2v − 1)2 = −4v(u + v)
On expanding using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, we get
u2 + u2 + 4v2 + 1 - 4uv - 2u + 4v = - 4vu - 4v2
2u2 + 4v2 - 2u + 4v + 4v2 + 1 = 0
On rearranging and taking common we get,
2(u2 - u) + 2 (4v2 + 2v) = - 1
Add values on both sides to complete the squares inside parenthesis
2(u2 – u + $$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?"

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