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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?

in Digital Logic by (355 points)
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getting 15.is it correct?

1 Answer

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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.
by Junior (759 points)
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How did gou got b =-26 and c=257???
0
how did u get 8??
+3

(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.

0
thanks @sushant  :)

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