In a quadratic equation whose coefficients are not necessarily real the constant term is not 0 The cube of the

let p and q are the roots. thus a/c

(p^2+q^2)^3 = (p^3+q^3)^2
solving both we get, p^2*q^2(3p^2+3q^2-2pq) = 0
now 3p^2+3q^2-2pq = 0.
p^2-(2/3*p*q)+q^2 = 0. thus here p and q are not real.
10 years agoHelpfull: Yes(10) No(3)

(b)is the correct option.bcz after calculating the values of the root we don't get any real roots.
10 years agoHelpfull: Yes(5) No(0)
i think 'a' is correct isn't it tanu.
10 years agoHelpfull: Yes(2) No(4)
Assume the given quadratic equation is ax2+bx+c=0 whose roots are p, q.
Now given that (α2+β2)3=(α3+β3)2
By expanding we get, α6+3.α4.β2+3.α2.β4+β6=α6+β6+2.α3.β3
3.α2.β2(α2+β2)=2.α3.β3
3.(α2+β2)=2.α.β
3.(α2+β2)+6.α.β−6.α.β=2.α.β
3.(α+β)2=8.α.β ...(1)
We know that sum of the roots = α+β=−ba
product of the roots = α.β=ca
Substituting in the equation (1) we get 3.(−ba)2=8.ca⇒3.b2=8.a.c

The nature of the roots can be determined by finding the magnitude of the determinant = b2−4ac
But we know that ac = 3b28
So b2−4ac = b2−4.3b28=−b88
9 years agoHelpfull: Yes(2) No(0)
@Tanu actually i hav solved and it is very large so i can't elobarte it. but finally i got the conclusion is that neither of them are real. and most important i want to know that how can u say that my ans is wrong.
10 years agoHelpfull: Yes(1) No(0)
@Tanu singh - Then solve it and Explain it.
10 years agoHelpfull: Yes(1) No(1)
@tanu..can u tell me the correct answer of it n explain its solution too..??
10 years agoHelpfull: Yes(1) No(0)
tanu if you know answer at-least say what is ans
10 years agoHelpfull: Yes(1) No(0)
Yes Tanu say atleast what is ans.
10 years agoHelpfull: Yes(1) No(0)
@AMIT The answer you gave is wrong but still please explain how reached on your conclusion?
10 years agoHelpfull: Yes(0) No(1)
@Amit Acually i know the correct option. I have answers ..i just need explanation.
10 years agoHelpfull: Yes(0) No(1)
@Ashutosh

As per your solution option b suits but its not correct.
10 years agoHelpfull: Yes(0) No(3)
@Everyone (c) is the correct option.
10 years agoHelpfull: Yes(0) No(0)
Roots are imaginary
8 years agoHelpfull: Yes(0) No(0)
Both root are imaginary so option 2 is correct
7 years agoHelpfull: Yes(0) No(0)

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