Math, asked by ayush33333sharma, 1 month ago

5
5.
If 2x2 + 3y2 + 4z2 – Voxy - 2/3yz - 22xz = 0,
then find the value of
(2x2 + 3y2 +1622 +276xy - 83yz - 8/2xz).​

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Answers

Answered by jitenderkumar0084
0

Answer:

Answer

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

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