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CAT 2023 Slot 1 — QA Question 5

Number TheoryEasy

If x and y are real numbers such that x2 + (x – 2y - 1)2 = -4y(x + y), then the value of x - 2y is?

Answer & solution

  • 1

  • B

    -1

  • C

    2

  • D

    0

Solution

Easy

Bring everything to one side and aim to write the left-hand expression as a sum of two perfect squares. A sum of squares of real numbers is zero only when each square is zero — that single observation cracks the whole problem.

1

Expand the right side and move it left:

x2+(x−2y−1)2=−4y(x+y)=−4xy−4y2x^2+(x-2y-1)^2=-4y(x+y)=-4xy-4y^2 ⇒ x2+4xy+4y2+(x−2y−1)2=0\Rightarrow\ x^2+4xy+4y^2+(x-2y-1)^2=0
2

Recognise the first three terms as a perfect square:

x2+4xy+4y2=(x+2y)2x^2+4xy+4y^2=(x+2y)^2 ⇒ (x+2y)2+(x−2y−1)2=0\Rightarrow\ (x+2y)^2+(x-2y-1)^2=0
3

Sum of two real squares =0=0 forces each to vanish:

x+2y=0andx−2y−1=0x+2y=0\qquad\text{and}\qquad x-2y-1=0
4

Read off the target. The second equation is exactly what we want:

x−2y=1x-2y=1
x−2y=1(option a)x-2y=\mathbf{1}\quad\text{(option a)}

You don't even need to solve for xx and yy separately. The instant the equation becomes (x+2y)2+(x−2y−1)2=0(x+2y)^2+(x-2y-1)^2=0, the second bracket must be 00, i.e. x−2y=1x-2y=1 directly.

CAT 2023 Slot 1 QA Q5: If x and y are real numbers such that x 2 + (x – 2y - 1) 2 = -4y(x + y), then the value of x - 2y is? — Solution | TheCATExam