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System of Quadratic Equations

CAT 2024 Slot 1 · Quantitative Ability · Medium · Quadratic Equations

This is a medium Quantitative Ability question from the CAT 2024 Slot 1 paper. It tests Quadratic Equations. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let x,y,x, y, and zz be real numbers satisfying 4(x2+y2+z2)=a4(x^2 + y^2 + z^2) = a 4(x−y−z)=3+a4(x - y - z) = 3 + a Then aa equals

  1. A.

    3

  2. B.

    1\frac{1}{3}

  3. C.

    1

  4. D.

    4

Answer

A

Explanation

Subtracting the first equation from the second: 4(x−x2)−4(y+y2)−4(z+z2)=34(x - x^2) - 4(y + y^2) - 4(z + z^2) = 3 1−(2x−1)2+1−(2y+1)2+1−(2z+1)2=31 - (2x - 1)^2 + 1 - (2y + 1)^2 + 1 - (2z + 1)^2 = 3 −(2x−1)2−(2y+1)2−(2z+1)2=0-(2x - 1)^2 - (2y + 1)^2 - (2z + 1)^2 = 0

Since squares of real numbers are non-negative, each term must be zero: x=12,y=−12,z=−12x = \frac{1}{2}, \quad y = -\frac{1}{2}, \quad z = -\frac{1}{2}

Substitute into a=4(x2+y2+z2)a = 4(x^2 + y^2 + z^2): a=4(14+14+14)=3a = 4\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\right) = 3

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