Question

if 9x^2+4y^2+z^2-6xy-2yz-3xz =0 ,then which of the following is true
1》2x=3y=z
2》3x=2y=z
3》2x=y=3z
4》3x=y=2z​

Answer 1

Answer:

2》3x=2y=z...............

Answer 2

$9x^{2}+4y^{2}+z^{2}-6xy-2yz-3xz=0\\=>(3x)^{2}+(2y)^{2}+(z)^{2}-(3x)(2y)-(2y)(z)-(3x)(z)=0$

Multiplying and dividing by two,

$=>\frac{1}{2}[18x^{2}+8y^{2}+2z^{2}-12xy-4yz-6xz]=0\\=>18x^{2}+8y^{2}+2z^{2}-12xy-4yz-6xz=0\\=>9x^{2}+4y^{2}-12xy+4y^{2}+z^{2}-4yz+z^{2}+9x^{2}-6xz=0\\=>(3x+2y)^{2}+(2y+z)^{2}+(z+3x)^{2}=0$

As the sum of 3 squares cannot be zero (squares are positive integers and their sums are 0 only when each of them is 0),

$3x+2y=0\\2y+z=0\\z+3x=0$

Solving each of them you get,

$3x=-2y\\2y=-z\\z=-3x\\=>3x=z\\2y=-z\\=>2y=-(-3x)\\=>2y=3x$

So, therefore, the answer is option (2) $3x=2y=z$

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