3 votes 3 votes In a particular number system the cubic equation X^3+bX^2+cX-190 has roots 5,8 and 9.What is the base of the no system? Digital Logic made-easy-test-series digital-logic number-system number-representation + – Gate Madrista asked Jan 22, 2017 retagged Mar 6, 2019 by adeebafatima1 Gate Madrista 2.5k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Sanket_ commented Jan 27, 2017 reply Follow Share getting 15.is it correct? 0 votes 0 votes Priyank Chauhan commented Feb 21, 2021 reply Follow Share How to solve this? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes X^3+bX^2+cX-190=(X-5)(X-8)(X-9) =X^3-(17+5)X^2+(72+85)X-360 NOW, b=-(17+5) and c=(72+85) take b=-26 and c=235 and let base is y. after solving in both case base is 8. Rajnish Kumar answered Jan 22, 2017 Rajnish Kumar comment Share Follow See all 4 Comments See all 4 4 Comments reply Rahul Jain25 commented Jan 22, 2017 reply Follow Share How did gou got b =-26 and c=257??? 0 votes 0 votes Akriti sood commented Jan 27, 2017 reply Follow Share how did u get 8?? 0 votes 0 votes Sushant Gokhale commented Jan 28, 2017 reply Follow Share (x-5)(x-8)(x-9) = 0 $\therefore$ x3 - 12x2 +67x -360 = 0 Now, comparing this equation with original, we get (190)y = (360)10 where, y is the new base $\therefore$ y2 + 9y = 360 $\therefore$ y=-24 or y=15 Discarding y=-24, we get y=15. 6 votes 6 votes Akriti sood commented Jan 28, 2017 reply Follow Share thanks @sushant :) 0 votes 0 votes Please log in or register to add a comment.