Algebraic Identity Optimization

CAT 2019 Slot 1 · QA · Medium · Linear and Quadratic Equations

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If u2+(u−2v−1)2=−4v(u+v)u^2 + (u - 2v - 1)^2 = -4v(u + v), then what is the value of u+3vu + 3v?

  1. A.

    14\frac{1}{4}

  2. B.

    12\frac{1}{2}

  3. C.

    0

  4. D.

    −14-\frac{1}{4}

Answer

D

Explanation

Expand equation: u2+u2+4v2+1−4uv−2u+4v=−4uv−4v2u^2 + u^2 + 4v^2 + 1 - 4uv - 2u + 4v = -4uv - 4v^2 2u2+8v2−2u+4v+1=02u^2 + 8v^2 - 2u + 4v + 1 = 0 2(u2−u+1/4)+8(v2+v/2+1/16)=02(u^2 - u + 1/4) + 8(v^2 + v/2 + 1/16) = 0 2(u−1/2)2+8(v+1/4)2=02(u - 1/2)^2 + 8(v + 1/4)^2 = 0 Since both terms are non-negative squares: u−1/2=0⇒u=1/2u - 1/2 = 0 \Rightarrow u = 1/2 v+1/4=0⇒v=−1/4v + 1/4 = 0 \Rightarrow v = -1/4 Then u+3v=1/2+3(−1/4)=1/2−3/4=−1/4u + 3v = 1/2 + 3(-1/4) = 1/2 - 3/4 = -1/4.

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