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In a quadratic equation, (whose coefficients are not necessarily real) the constant term is not 0. The cube of the sum of the squares of its roots is equal to the square of the sum of the cubes of its roots. Which of the following is true?
option
a) Both roots are real
b) Neither of the roots is real
c) At least one root is non-real
d) At least one root is real
Read Solution (Total 15)
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- 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. - 11 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.
- 11 years agoHelpfull: Yes(5) No(0)
- i think 'a' is correct isn't it tanu.
- 11 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 - 10 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.
- 11 years agoHelpfull: Yes(1) No(0)
- @Tanu singh - Then solve it and Explain it.
- 11 years agoHelpfull: Yes(1) No(1)
- @tanu..can u tell me the correct answer of it n explain its solution too..??
- 11 years agoHelpfull: Yes(1) No(0)
- tanu if you know answer at-least say what is ans
- 11 years agoHelpfull: Yes(1) No(0)
- Yes Tanu say atleast what is ans.
- 11 years agoHelpfull: Yes(1) No(0)
- @AMIT The answer you gave is wrong but still please explain how reached on your conclusion?
- 11 years agoHelpfull: Yes(0) No(1)
- @Amit Acually i know the correct option. I have answers ..i just need explanation.
- 11 years agoHelpfull: Yes(0) No(1)
- @Ashutosh
As per your solution option b suits but its not correct. - 11 years agoHelpfull: Yes(0) No(3)
- @Everyone (c) is the correct option.
- 11 years agoHelpfull: Yes(0) No(0)
- Roots are imaginary
- 9 years agoHelpfull: Yes(0) No(0)
- Both root are imaginary so option 2 is correct
- 8 years agoHelpfull: Yes(0) No(0)
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