SOLUTION: P(X)=24X^3-26X^2+29X-6 HAS THREE REAL ROOTS ONE OF THEM IS 2/3 THE OTHERS ARE ____ AND ____

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Question 18127: P(X)=24X^3-26X^2+29X-6 HAS THREE REAL ROOTS ONE OF THEM IS 2/3 THE OTHERS ARE ____ AND ____
Answer by venugopalramana(3286) About Me  (Show Source):
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SEE THE FOLLOWING AND DO IN A SIMILAR WAY
P(X)=3X^3-10X^2-5X+2 HAS THREE REAL ROOTS ONE OF THEM IS -2/3 THE OTHERS ARE ____ AND ____
one root is -2/3.let the other 2 roots be x,y
sum of roots =x+y-2/3=10/3
so x+y=10/3+2/3=12/3=4......x+y=4...........(1)
product of roots x*y*(-2/3)=-(2/3)
hence xy = 2*3/(3*2)=1
(x-y)^2=(x+y)^2-4xy=4^2-4*1=16-4=12
x-y=sqrt12=2sqrt3.................(2)
(1)+(2).....2x=4+2sqrt3...or x=2+sqrt5
(1)-(2).....2y=4-2sqrt5......y=2-sqrt3