The question is attached above

Asked by sudhanshubhushanroy | 19th Nov, 2017, 09:10: PM

Expert Answer:

begin mathsize 16px style straight x squared plus straight b subscript 2 straight x plus straight c subscript 2 equals 0
rightwards double arrow straight b subscript 2 squared minus 4 straight c subscript 2 greater than 0
rightwards double arrow straight b subscript 2 squared greater or equal than 4 straight c subscript 2
Given space that space
straight b subscript 1 straight b subscript 2 equals 2 open parentheses straight c subscript 1 plus straight c subscript 2 close parentheses
straight b subscript 1 squared straight b subscript 2 squared equals 4 open parentheses straight c subscript 1 plus straight c subscript 2 close parentheses squared
straight b subscript 1 squared straight b subscript 2 squared equals 4 open parentheses straight c subscript 1 minus straight c subscript 2 close parentheses squared plus 16 straight c subscript 1 straight c subscript 2
straight b subscript 1 squared straight b subscript 2 squared minus 16 straight c subscript 1 straight c subscript 2 equals 4 open parentheses straight c subscript 1 minus straight c subscript 2 close parentheses squared
As space 4 open parentheses straight c subscript 1 minus straight c subscript 2 close parentheses squared greater or equal than 0
rightwards double arrow straight b subscript 1 squared straight b subscript 2 squared minus 16 straight c subscript 1 straight c subscript 2 greater or equal than 0
Hence comma space it space has space real space roots. end style

Answered by Sneha shidid | 30th Nov, 2017, 10:42: AM

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