Math, asked by sandhyasappa205, 1 month ago


If x square+ 4y square + 9z square + 6yz - 3zx - 2xy = 0, then​

Answers

Answered by kapilchavhan223
29

Step-by-step explanation:

x^2+4y^2+9z^2–2xy-6yz-3xz=0

or x^2–2xy+4y^2–6yz+9z^2–3xz=0

or x(x-2y)+2y(2y-3z)+3z(3z-x)=0

Therefore x-2y=0

x=2y or x/2=y/1=k(let)

x=2k , y=k ,

2y-3z=0 put y=k

2k-3z=0 or z=(2/3)k

Now x : y : z = 2k : k : 2k /3

x : y : z = 2 : 1 : 2/3

x : y : z =2×3 : 1×3 : ( 2/3)×3

x : y : z = 6 : 3 : 2 .

Answered by Anonymous
2

\huge \colorbox{cyan}{ᴀɴsᴡᴇʀ}

x^2+4y^2+9z^2–2xy-6yz-3xz=0

or x^2–2xy+4y^2–6yz+9z^2–3xz=0

or x(x-2y)+2y(2y-3z)+3z(3z-x)=0

Therefore x-2y=0

x=2y or x/2=y/1=k(let)

x=2k , y=k ,

2y-3z=0 put y=k

2k-3z=0 or z=(2/3)k

Now x : y : z = 2k : k : 2k /3

x : y : z = 2 : 1 : 2/3

x : y : z =2×3 : 1×3 : ( 2/3)×3

x : y : z = 6 : 3 : 2 .

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