CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
Vikramjeet Singh Virk

· started a discussion

· 1 Months ago

short tricks (Question: 214494.0 )

provide the videos with tricky solutions that save time during the exam

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:

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
WE have inserted video, please go through it.
Keep Learning with Us
Best Wishes
Team Toprankers
-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions