1heyyyyyyyyy............ koi toh rep karoooooooooooo!!!!!!!!!!!!!!!!
For how many values of "a" will the following equation have two equal roots:
x^3 + (a^2++a-1)x^2 + (a^3-a^2-a)x -a^3 =0
i got 4 as ans....
plz verify...
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15 Answers
1let the roots be x.
Then
3x = -(a2 + a - 1)
3x2 = (a3 - a2 - a)
x3 = -a3 = > (x=-a)
Therefore substituting x=-a in the first two equation you get 4 solutions.
So i think 4 is correct
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21@sky : is a given to be real?????
this un's a nice questn.....[1]
I wud like 2 solve it....
hope that "++" in the coeff of x^2 is same as "+"
341I am assuming that a is real.
Notice that x = -a is a solution.
a= 0 is admissible (with x=0 is being the repeated root)
Now, if a is non-zero we have two possibilities:
(1) -a is the repeated root.
(2) -a is not repeated.
Case 1: (-a) is repeated. Then since product of roots is a3, the third root is a.
Sum of roots = (-a)+(-a)+a = -a = 1-a-a2 means a =±1 and it is easily verified that both values of a are admissible
Case (2). -a is not repeated.
In that case the other root is ±ai
But that is impossible as we know that if ai is a root, -ai must also be (since the coefficients are real, the non-real roots come in conjugate pairs). Hence a must be zero, but we have already assumed a non-zero.
So we have three possibilities a=0 , 1 and -1
(where have I missed the 4th one?)
11skygirl dont worry about these type of ques as in jee syllabus only quad eq of degree 2 is included
but acc to my knowledge 1 root of 3 degree eq has to be found out by hit and and trial
and the remaining 2 roots are found with the help of that 1 root
1oh sry bhaiya...
main yeh question ke baare me bhool hi gayi thi ...
its long tym back...
sry [2]
jus replying ...
