Question

if 4x²+y²+4z²+3=4z+2y-4z, then the value of x+y+z is:
a) 1
b)3/2
c)2
d)0​

Answer 1

Answer:

a) 1

Step-by-step explanation:

4x^2 + y^2 + 4z^2 + 3 - 4x -2y + 4z = 0

=> (1-2x)^2 + (1-y)^2 + (1+2z)^2 =0

=> x=1/2, y =1, z=-1/2

=> x+y+z = 1.

Answer 2

$answer$

$4 {x}^{2} + {y}^{2} + 4 {z}^{2} + 3 - 4x - 2y + 4z = 0$

$= > {(1 - 2x)}^{2} + {( 1- y)}^{2} + {1 + 2z}^{2} = 0$

$= > x = \frac{1}{2} \: \: y = 1 \: \: z = \frac{ - 1}{2}$

$x + y + z = 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1$

$hope \: this \: helps \: u.....$