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CAT 2018 Slot 1 — QA Question 25

Solving Quadratic EquationsEasy

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

Answer & solution

Correct answer: -1/4

  • A

    1/2

  • -1/4

  • C

    0

  • D

    1/4

Solution

Easy

Bring everything to one side and try to write the expression as a sum of squares. A sum of squares equal to zero forces each square to be zero, which pins down uu and vv uniquely.

1

Move all terms to the left and expand the −4v(u+v)-4v(u+v) piece.

u2+(u−2v−1)2=−4v(u+v)⇒ u2+(u−2v−1)2+4uv+4v2=0(bring RHS over)⇒ (u−2v−1)2+(u2+4uv+4v2)=0\begin{aligned} &u^2+(u-2v-1)^2 = -4v(u+v)\\ &\Rightarrow\ u^2+(u-2v-1)^2+4uv+4v^2 = 0 \quad\text{(bring RHS over)}\\ &\Rightarrow\ (u-2v-1)^2+\big(u^2+4uv+4v^2\big)=0 \end{aligned}
2

Recognise the perfect square. Note u2+4uv+4v2=(u+2v)2u^2+4uv+4v^2=(u+2v)^2.

(u−2v−1)2+(u+2v)2=0\begin{aligned} &(u-2v-1)^2+(u+2v)^2 = 0 \end{aligned}

Both terms are non-negative, so a sum equal to zero forces each to be zero.

3

Set each square to zero and solve the pair.

u−2v−1=0(first square=0)u+2v=0(second square=0)⇒ 2u−1=0(add the two equations)⇒ u=12,v=−u2=−14\begin{aligned} &u-2v-1 = 0 \quad\text{(first square}=0)\\ &u+2v = 0 \quad\text{(second square}=0)\\ &\Rightarrow\ 2u-1 = 0 \quad\text{(add the two equations)}\\ &\Rightarrow\ u = \tfrac{1}{2},\qquad v = -\tfrac{u}{2} = -\tfrac{1}{4} \end{aligned}
4

Compute u+3vu+3v.

u+3v=12+3 ⁣(−14)=12−34=−14u+3v = \tfrac{1}{2}+3\!\left(-\tfrac{1}{4}\right) = \tfrac{1}{2}-\tfrac{3}{4} = -\tfrac{1}{4}
u+3v=−14u+3v = -\tfrac{1}{4}

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CAT 2018 Slot 1 QA Q25 — Solving Quadratic Equations solution | TheCATExam