Question

Let \(x, y, z\) be real numbers such that \(4(x^2 + y^2 + z^2) = a\) and \(4(x - y - z) = 3 + a\), then find the value of \(a\).

• A. 1
• B. 1 and \(\frac{1}{3}\)
• C. 4
• D. 3

Answer 1

The Correct Option is D

Given the equations: 

\(4(x^2+y^2+z^2)=a\) (1)
\(4(x-y-z)=3+a\) (2)

We need to find the value of \(a\).

First, let's solve equation (2) for \(x\):
\(4x - 4y - 4z = 3 + a\)
Rearranging gives:
\(4x = 3 + a + 4y + 4z\)
\(x = \frac{3+a+4y+4z}{4}\) (3)

Now substitute \(x\) from equation (3) into equation (1):
\(4\left(\left(\frac{3+a+4y+4z}{4}\right)^2 + y^2 + z^2\right) = a\)

Expanding and simplifying:
\(\left(\frac{3+a+4y+4z}{4}\right)^2 = \frac{(3+a+4y+4z)^2}{16}\)
Thus, equation (1) becomes:
\(4\left(\frac{(3+a+4y+4z)^2}{16} + y^2 + z^2\right) = a\)

Simplify the left-hand side:
\(\frac{(3+a+4y+4z)^2}{4} + 4(y^2 + z^2) = a\)

We know from (1):
\(4(y^2+z^2)=a-x^2\)
Substitute:
\(\frac{(3+a+4y+4z)^2}{4} + a-x^2 = a\)

Let \(x=y=z\):
\(a=4(3x^2)\Rightarrow a=12x^2\)
From (2):
\(4(0)=3+a \Rightarrow a=3\)

Thus, the value of \(a\) is: \(3\).