(xiv) if x ²+5y² +9z^2-2y(2x + 3z)=0 so find the value of x:y:z = ?
Answers
Answer:
x:y:z = 6:3:1
Step-by-step explanation:
The given equation is
{x}^{2} + 5 {y}^{2} + 9 {z}^{2} - 2y(2x + 3z) = 0x
2
+5y
2
+9z
2
−2y(2x+3z)=0
{x}^{2} + 5 {y}^{2} + 9 {z}^{2} - 4xy - 6xz = 0x
2
+5y
2
+9z
2
−4xy−6xz=0
( {x}^{2} - 4xy + 4 {y}^{2} ) + ( {y}^{2} - 6xz + 9 {z}^{2} ) = 0(x
2
−4xy+4y
2
)+(y
2
−6xz+9z
2
)=0
{(x - 2y)}^{2} + {(y - 3z)}^{2} = 0(x−2y)
2
+(y−3z)
2
=0
since \: \: {(x - 2y)}^{2} \geqslant 0since(x−2y)
2
⩾0
and \: {(y - 3z)}^{2} \geqslant 0and(y−3z)
2
⩾0
the above equation is possible only if both terms are zero.
{(x - 2y)}^{2} =\: {(y - 3z)}^{2} = 0(x−2y)
2
=(y−3z)
2
=0
(x - 2y) = (y - 3z) = 0(x−2y)=(y−3z)=0
x = 2y \: \: and \: y = 3zx=2yandy=3z
\frac{x}{2} = y = 3z
2
x
=y=3z
\frac{x}{6} = \frac{y}{3} = \frac{z}{1}
6
x
=
3
y
=
1
z
Hence x:y:z = 6:3:1please mark me brainlist