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Home / Doubts and Solutions/JEE/Mathematics

Q) Solve the equation x^3-11x^2+36x-36=0, if its roots are in HP.

Asked by Anish 4th September 2018, 4:30 PM
Answered by Expert
Answer:
begin mathsize 18px style straight x cubed minus 11 straight x squared plus 36 straight x minus 36 equals 0 let space the space roots space be space straight a comma straight b space and space straight c straight x cubed minus 11 straight x squared plus 36 straight x minus 36 equals open parentheses straight x minus straight a close parentheses open parentheses straight x minus straight b close parentheses open parentheses straight x minus straight c close parentheses open parentheses straight x minus straight a close parentheses open parentheses straight x minus straight b close parentheses open parentheses straight x minus straight c close parentheses equals straight x cubed minus open parentheses straight a plus straight b plus straight c close parentheses straight x squared plus open parentheses ab plus ac plus bc close parentheses straight x minus abc comparing space with space straight x cubed minus 11 straight x squared plus 36 straight x minus 36 equals 0 we space get straight a plus straight b plus straight c equals 11.....1 ab plus ac plus bc equals 36.......2 abc equals 36.......3 also space we space know 2 over straight b equals 1 over straight a plus 1 over straight c 2 over straight b equals space fraction numerator straight a plus straight c over denominator ac end fraction fraction numerator 2 ac over denominator straight a plus straight c end fraction space equals space straight b eqn 2 divided by eqn 3 1 over straight c plus 1 over straight a plus 1 over straight b equals 1 2 over straight b plus 1 over straight b equals 1 straight b equals 3 straight a plus straight c equals 8 ac space equals 12 straight a space equals space 2 space and space straight c space equals 6 roots space are space 2 comma 3 comma 6 end style
Answered by Expert 6th September 2018, 11:52 AM
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