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CBSE Class 11-science Answered

Find the condition when one root is equal to the square of other if  the equation is of the form: ax^2+bx+c = 0 .
Asked by Sanjay | 06 Jun, 2015, 06:43: PM
answered-by-expert Expert Answer
C o n s i d e r space t h e space e q u a t i o n space a x squared plus b x plus c equals 0 G i v e n space t h a t space o n e space r o o t space o f space t h e space a b o v e space e q u a t i o n space i s space e q u a l space t o space t h e space s q u a r e space o f space t h e space o t h e r. L e t space t h e space r o o t s space o f space t h e space g i v e n space e q u a t i o n space b e space alpha space a n d space alpha squared. W e space k n o w space t h a t comma S u m space o f space t h e space r o o t s equals fraction numerator negative b over denominator a end fraction P r o d u c t space o f space t h e space r o o t s equals c over a T h u s comma space w e space h a v e comma alpha plus space alpha squared equals fraction numerator negative b over denominator a end fraction... left parenthesis 1 right parenthesis a n d alpha cross times alpha squared equals c over a rightwards double arrow alpha cubed equals open parentheses c over a close parentheses rightwards double arrow alpha equals open parentheses c over a close parentheses to the power of 1 third end exponent... left parenthesis 2 right parenthesis S u b s t i t u t i n g space t h e space v a l u e space o f space alpha space f r o m space e q u a t i o n space left parenthesis 2 right parenthesis space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e comma open parentheses c over a close parentheses to the power of 1 third end exponent plus space open parentheses c over a close parentheses to the power of 2 over 3 end exponent equals fraction numerator negative b over denominator a end fraction.. left parenthesis 3 right parenthesis E q u a t i o n space left parenthesis 3 right parenthesis space i s space t h e space r e q u i r e d space c o n d i t i o n.
Answered by Vimala Ramamurthy | 07 Jun, 2015, 02:58: PM

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