IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

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IF THE QUADRATIC EQN 3X2+PX+3 HAS ROOTS;a and a2 then find the value of p such that p>0.

johri_anshuman

product of the roots=a*a^2=a^3=1
=>a=1

so the equation has repeated roots at x=1

as the equation has repeated roots so D=0
=>p^2-4.3.3=0
=>p^2=36
=>p=+-6

as p>0
so p=6

nivedh_89

a+a^2 = -p/3.....
a^3 = 1........therefore...a = 1.....
2 = -p/3....
p = -6.........but......
determinant = 0...therefore p = 6.......

krishna.gopal

If a³ = 1 then either two roots are 1,1 (therefore p = -6) or they may also be

,

² where

is imaginary cube root of unity. For second case sum of roots =-1 which means p = 3.

So for p = 3 is only case when p>0 is possible

spideyunlimited

product of roots
a^3 = 1
therefore a = 1, w, w^2
also we know that 1 + w + w^2 = 0

sum of roots
a^2 + a = -p/3
substitute a^2 by 1/ a ( as acube = 1) and we get

(1+ a^2 ) / a = - p/3

as p>0
left hand side can only be -ve when a = w out of 1 , w and w^2
thus
a^2 + a = - p/3
==>> w^2 + w = -p/3
- 1 = -p/3 .. (as w^2 + w + 1 = 0)

p = 3

PLZ RATE :)

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