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Siddhesh Mishra

· started a discussion

· 1 Months ago

Wrong Question (Question:646949.0)

x+(1/6) should be written as x+(1/x)

Question:
If 4a² + 9b² + 25c² = 6ab + 15bc + 10ac, then the relation among a, b, c is:
Options:
A) 2a + 3b + 5c = 0 
B) a = b = c
C) a = \(\cfrac{3b}{2}\) and b = \(\cfrac{5c}{3}\).
D) 2a + 3b = 5c
Solution:
Ans: (c) It is given that  

⇒ 4a² + 9b² + 25c² = 6ab + 15bc + 10ac

⇒ (2a)² + (3b)² + (5c)² – (2a)(3b) – (3b)(5c) – (5c)(2a) = 0

⇒ \(\cfrac{1}{2}\) [(2a – 3b)² + (3b – 5c)² + (5c – 2a)²] = 0

⇒ 2a = 3b = 5c

⇒ a = \(\cfrac{3b}{2}\) and b = \(\cfrac{5c}{3}\).

Knowledge Expert

· commented

· 1 Months ago

Dear student
Please specify your doubt clearly.
There is no variable (x) in the given question.

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