find 'a' ....

???

heyyyyyyyyy............ koi toh rep karoooooooooooo!!!!!!!!!!!!!!!!

let 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

Lol wait i thought sky meant 4 sols... I will slove -.-

I 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?)

no reactions?

skygirl 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

kk...

i din get your soln fully ...

Which part did you get stuck at?

So whats the final ans ???3 or 4 ??