Question

If 2x^2 + 3y^2 + 4z^2 - √6xy - 2√3yz-2√2xz = 0,then find the value of
(2x^2 + 3y^2 +16z^² +2√6xy - 8√3y2- 8√2xz).​

Answer 1

Step-by-step explanation:

If 2x^2 + 3y^2 + 4z^2 - √6xy - 2√3yz-2√2xz = 0,

then find the value of

(2x^2 + 3y^2 +16z^² +2√6xy - 8√3yz- 8√2xz).

2x² + 3y² + 4z² - √6xy - 2√3yz-2√2xz = 0

multiplying by 2

=> 4x² + 6y² + 8z² - 2√6xy - 4√3yz-4√2xz = 0

=> 4x² + 6y² + 8z² - 2√6xy - 4√3yz-4√2xz = 0

=> 2x² + 2x² + 3y² + 3y² + 4z²  + 4z²- 2√6xy - 4√3yz-4√2xz = 0

=> 2x² + 3y²- 2√6xy + 2x² + 4z² + -4√2xz + 3y² + 4z² - 4√3yz = 0

=> ( √2x - √3y)² + (√2x - 2z)² + (√3y - 2z)² = 0

As square can not be negative

=> (√2x - √3y)² = (√2x - 2z)² = (√3y - 2z)² = 0

=>  √2x = √3y  , √2x = 2z , √3y = 2z

=> √2x = √3y = 2z

=> 2x² = 3y² = 4z²

2x² + 3y² + 16z² + 2√6xy - 8√3yz - 8√2xz

Putting values of y & z in form of x

= 2x² + 2x² + 8x² + 2√2x√2x -  8√2x.x/√2  - 8√2x.x/√2

= 12x² + 4x² - 8x² - 8x²

= 16x² - 16x²

= 0

2x² + 3y² + 16z² + 2√6xy - 8√3yz - 8√2xz = 0