Question

Let $x, y,$ and $z$ be real numbers satisfying $$4(x^2 + y^2 + z^2) = a$$ $$4(x - y - z) = 3 + a$$ Then $a$ equals

Accepted Answer

A. Subtracting the first equation from the second: $$4(x - x^2) - 4(y + y^2) - 4(z + z^2) = 3$$ $$1 - (2x - 1)^2 + 1 - (2y + 1)^2 + 1 - (2z + 1)^2 = 3$$ $$-(2x - 1)^2 - (2y + 1)^2 - (2z + 1)^2 = 0$$ Since squares of real numbers are non-negative, each term must be zero: $$x = \frac{1}{2}, \quad y = -\frac{1}{2}, \quad z = -\frac{1}{2}$$ Substitute into $a = 4(x^2 + y^2 + z^2)$: $$a = 4\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\right) = 3$$

System of Quadratic Equations

Explanation

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