Question

यदि 9 a ^ 2 + 4b ^ 2 + c ^ 2 + 21 = 4 (3a + b-2c) तो (9a + 4b-c) का मान है​

Answer 1

9a + 4b-c = 12  if 9a²  + 4b²  + c²  + 21 = 4 (3a + b - 2c)

Step-by-step explanation:

9a²  + 4b²  + c²  + 21 = 4 (3a + b - 2c)

= 9a² - 12a  + 4b² - 4b + c² + 8c + 21 = 0

=> (3a - 2)² - 4  + (2b - 1)² - 1  + (c + 4)²  - 16  + 21 = 0

=>  (3a - 2)² +  (2b - 1)² + (c + 4)² = 0

=> a = 2/3  

   b = 1/2

  c = -4

9a + 4b - c  = 9(2/3)  + 4(1/2) - (-4)

= 6 + 2 + 4

= 12

Learn more:

4a^2+9b^2+16c^2+1/9a^2+1/16b^2+25c^2=133/30.

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If 9a^2 + 16b^2 + c^2 + 25 = 24 (a + b) then find 3a+4b+5c.​

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