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If a, b are the roots of x2 + px + q = 0, and w is a cube root of unity, then value of  (ωα + ω2β)(ω2α + ωβ) is
  • a)
    p2
  • b)
    3q
  • c)
    p2 – 2q
  • d)
    p2 – 3q 
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If a, b are the roots of x2 + px + q = 0, and w is a cube root of unit...
We have, α + β = -p,αβ = q
Now (ωα + ω2β)(ω2α + ωβ) = ω3α24αβ+ω2αβ + ω3β2
= α22 +(ω+ω2)αβ = α22 -αβ = (α + β)2 -3αβ = p2 -3q
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Most Upvoted Answer
If a, b are the roots of x2 + px + q = 0, and w is a cube root of unit...
Understanding the Problem
Given the quadratic equation \(x^2 + px + q = 0\), let \(a\) and \(b\) be its roots. By Vieta's formulas, we have:
- Sum of roots: \(a + b = -p\)
- Product of roots: \(ab = q\)
Let \(\omega\) be a primitive cube root of unity, satisfying \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).
Expression Breakdown
We need to evaluate the expression \((\omega a + \omega^2 b)(\omega^2 a + \omega b)\).
1. Expanding the Expression:
\[
(\omega a + \omega^2 b)(\omega^2 a + \omega b) = \omega^3 a^2 + \omega^2 ab + \omega a b + \omega^3 b^2
\]
Since \(\omega^3 = 1\), this simplifies to:
\[
a^2 + b^2 + (\omega^2 + \omega)ab = a^2 + b^2 - ab
\]
2. Calculating \(a^2 + b^2\):
Using the identity \(a^2 + b^2 = (a + b)^2 - 2ab\):
\[
a^2 + b^2 = (-p)^2 - 2q = p^2 - 2q
\]
3. Combining Results:
Substitute \(a^2 + b^2\) back into our expression:
\[
a^2 + b^2 - ab = (p^2 - 2q) - q = p^2 - 3q
\]
Final Result
Thus, the value of \((\omega a + \omega^2 b)(\omega^2 a + \omega b)\) is:
\[
\boxed{p^2 - 3q}
\]
This confirms that the correct option is \(d) p^2 - 3q\).
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If a, b are the roots of x2 + px + q = 0, and w is a cube root of unity, then value of (ωα+ω2β)(ω2α+ωβ) isa)p2b)3qc)p2 – 2qd)p2 – 3qCorrect answer is option 'D'. Can you explain this answer?
Question Description
If a, b are the roots of x2 + px + q = 0, and w is a cube root of unity, then value of (ωα+ω2β)(ω2α+ωβ) isa)p2b)3qc)p2 – 2qd)p2 – 3qCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a, b are the roots of x2 + px + q = 0, and w is a cube root of unity, then value of (ωα+ω2β)(ω2α+ωβ) isa)p2b)3qc)p2 – 2qd)p2 – 3qCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a, b are the roots of x2 + px + q = 0, and w is a cube root of unity, then value of (ωα+ω2β)(ω2α+ωβ) isa)p2b)3qc)p2 – 2qd)p2 – 3qCorrect answer is option 'D'. Can you explain this answer?.
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