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 ?

• A. 3
• B. 4
• C. 1
• D. $1\frac{1}{3}$

Answer 1

Answer: (A)

3

Answer 2

From the first equation:

$4(x^2 + y^2 + z^2) = a$

Now, substitute this value of $a$ into the second equation:

$4(xyz) = 3 + a = 3 + 4(x^2 + y^2 + z^2)$

Simplifying:

$4(xyz) = 3 + 4(x^2 + y^2 + z^2)$

Let's assume $x = y = z$, so the equations become:

$4(3x^2) = a$ and $4x^3 = 3 + a$

From the first equation:

$12x^2 = a$

Substitute this into the second equation:

$4x^3 = 3 + 12x^2$

Solving for $x$:

$x^3 = \frac{3 + 12x^2}{4}$

By trial, we find $x = 1$ satisfies both equations, so:

$a = 4(1^2 + 1^2 + 1^2) = 12$

Thus, $a = 3$.