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Sandeep kumar

· started a discussion

· 1 Months ago

Totally useless solutions (Question: 214494.0 )

Very disappointed by toprankers, definitely not a sure shot series. The solutions provided are useless

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 = \(\cfrac{5c}{3}\).

For Video Solution Click on the link given below:

Talented Boy

· commented

· 1 Months ago

You must know the algebric identities , then you will be able to understand the solution. Don't jump to fire words.
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