Let x, y, and z be real numbers satisfying 4(x2+y2+z2)=a4(x−y−z)=3+a Then a equals
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Official Correct Answer: B. The question says:
Let x, y, and z be real numbers satisfying
4(x² + y² + z²) = a
4(x - y - z) = 3 + a
Then a equals...
And the options are:
A) 1 1/3
B) 3
C) 1
D) 4
Alright, so we have two equations with three variables x, y, z, and we need to find the value of 'a'. Hmm, that seems a bit tricky because usually, with two equations and three variables, we can't solve for all variables uniquely. But maybe there's a way to relate them or find a relationship that allows us to find 'a' without knowing x, y, z individually.
We have the equations:
1) 4(x² + y² + z²) = a
2) 4(x - y - z) = 3 + a
We find 'a'. if I can manipulate these equations to eliminate x, y, z or find a relationship between them.
First, equation 2 seems simpler, so maybe I can express (x - y - z) in terms of 'a'. Let me do that.
From equation 2:
4(x - y - z) = 3 + a
Divide both sides by 4:
x - y - z = (3 + a)/4
Let this = equation 3:
3) x - y - z = (3 + a)/4
Now, equation 1 is about the sum of squares. Maybe I can relate this to equation 3. for real numbers, the square of a sum is equal to the sum of squares plus twice the sum of the products. But here, I have a linear combination in equation 3.
Wait, maybe I can think of (x - y - z) as a single variable. Let S = x - y - z. Then equation 3 becomes S = (3 + a)/4.
But how does that help me with equation 1? Equation 1 is 4(x² + y² + z²) = a. Hmm.
I wonder if I can express x² + y² + z² in terms of S and maybe some other terms. Alternatively, maybe I can use the Cauchy-Schwarz inequality or some identity. Wait, another idea: perhaps express x in terms of y and z from equation 3 and substitute into equation 1.
From equation 3:
x = y + z + (3 + a)/4
Substituting this into equation 1.
So, equation 1 becomes:
4[(y + z + (3 + a)/4)² + y² + z²] = a
Expanding this step by step.
First, Compute (y + z + (3 + a)/4)².
Let A = y + z, and B = (3 + a)/4. Then, (A + B)² = A² + 2AB + B².
So, expanding (y + z + (3 + a)/4)²:
= (y + z)² + 2(y + z)((3 + a)/4) + ((3 + a)/4)²
= y² + 2yz + z² + (2*(3 + a)/4)(y + z) + (3 + a)² / 16
Simplify the middle term:
2*(3 + a)/4 = (3 + a)/2
So, putting it all together:
= y² + 2yz + z² + (3 + a)/2 * (y + z) + (3 + a)² / 16
Now, substitute this back into equation 1:
4[ (y² + 2yz + z² + (3 + a)/2*(y + z) + (3 + a)² / 16 ) + y² + z² ] = a
Simplifying inside the brackets first:
First, combine like terms:
= y² + 2yz + z² + (3 + a)/2*(y + z) + (3 + a)² / 16 + y² + z²
= (y² + y²) + (z² + z²) + 2yz + (3 + a)/2*(y + z) + (3 + a)² / 16
= 2y² + 2z² + 2yz + (3 + a)/2*(y + z) + (3 + a)² / 16
So, equation 1 becomes:
4[2y² + 2z² + 2yz + (3 + a)/2*(y + z) + (3 + a)² / 16] = a
Factoring out the 4:
4*2y² + 4*2z² + 4*2yz + 4*(3 + a)/2*(y + z) + 4*(3 + a)² / 16 = a
Simplify each term:
= 8y² + 8z² + 8yz + 2*(3 + a)*(y + z) + (3 + a)² / 4 = a
Hmm, this is getting a bit complicated. Maybe there's a better approach.
Wait, another idea: Let me consider that x, y, z are real numbers, so the expression x² + y² + z² must be non-negative. Also, from equation 1, 4(x² + y² + z²) = a, so a must be non-negative as well.
Looking at the options, all are positive, so that's consistent.
But how else can I relate equation 1 and equation 2?
Wait, perhaps I can use the identity:
(x - y - z)² = x² + y² + z² - 2xy - 2xz + 2yz
But I'm not sure if that helps directly. Alternatively, maybe I can think of (x - y - z) as a vector and relate it to the norm.
Alternatively, perhaps I can use the Cauchy-Schwarz inequality, which states that (x - y - z)^2 ≤ (1^2 + (-1)^2 + (-1)^2)(x² + y² + z²).
Yes, Cauchy-Schwarz inequality says that (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v), where u and v are vectors.
In this case, let me take u = (1, -1, -1) and v = (x, y, z). Then, u ⋅ v = x - y - z, which is equal to (3 + a)/4 from equation 3