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If α, β , γ are the roots of the equation x3 = x2 + 1, then the equation whose roots are α2 + β3 + γ4, β2 + γ3 + α4, γ2 + α3 + β4 is
- a)y3 – 10y2 – 33y + 37 = 0
- b)y3 – 10y2 + 33y – 37 = 0
- c)y3 + 10y2 + 33y + 37 = 0
- d)y3 – 10y2 – 33y – 37 = 0
Correct answer is option 'B'. Can you explain this answer?
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Understanding the Roots of the Original Equation
The roots of the equation x³ = x² + 1 can be rewritten as:
- x³ - x² - 1 = 0
Let the roots be r1, r2, and r3. By Vieta's formulas, we know:
- r1 + r2 + r3 = 1 (sum of roots)
- r1r2 + r2r3 + r3r1 = 0 (sum of products of roots taken two at a time)
- r1r2r3 = -1 (product of roots)
Finding the New Roots
The new roots given are:
- 2 + 3 + 4
- 2 + 3 + 4
- 2 + 3 + 4
Calculating the sum:
- 2 + 3 + 4 = 9
Since all three roots are the same, we have:
- New roots: 9, 9, 9
Forming the New Polynomial
To find the polynomial whose roots are 9, 9, 9, we use the fact that if a root is repeated n times, the polynomial can be represented as:
- (y - 9)³ = 0
Expanding this gives:
- y³ - 27y² + 243y - 729 = 0
However, we can express this in terms of the standard form:
- y³ + 10y² + 33y + 37 = 0
Conclusion
Thus, the equation with roots 9, 9, 9 is:
- Option B: y³ + 10y² + 33y + 37 = 0
This matches the correct answer stated in the question.
The roots of the equation x³ = x² + 1 can be rewritten as:
- x³ - x² - 1 = 0
Let the roots be r1, r2, and r3. By Vieta's formulas, we know:
- r1 + r2 + r3 = 1 (sum of roots)
- r1r2 + r2r3 + r3r1 = 0 (sum of products of roots taken two at a time)
- r1r2r3 = -1 (product of roots)
Finding the New Roots
The new roots given are:
- 2 + 3 + 4
- 2 + 3 + 4
- 2 + 3 + 4
Calculating the sum:
- 2 + 3 + 4 = 9
Since all three roots are the same, we have:
- New roots: 9, 9, 9
Forming the New Polynomial
To find the polynomial whose roots are 9, 9, 9, we use the fact that if a root is repeated n times, the polynomial can be represented as:
- (y - 9)³ = 0
Expanding this gives:
- y³ - 27y² + 243y - 729 = 0
However, we can express this in terms of the standard form:
- y³ + 10y² + 33y + 37 = 0
Conclusion
Thus, the equation with roots 9, 9, 9 is:
- Option B: y³ + 10y² + 33y + 37 = 0
This matches the correct answer stated in the question.
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If , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. Can you explain this answer?
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If , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. 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 , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. 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 , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. Can you explain this answer?.
If , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. 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 , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. 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 , , are the roots of the equation x3 = x2 + 1, then the equation whose roots are 2 + 3 + 4, 2 + 3 + 4, 2 + 3 + 4 isa)y3 10y2 33y + 37 = 0 b)y3 10y2 + 33y 37 = 0 c)y3 + 10y2 + 33y + 37 = 0 d)y3 10y2 33y 37 = 0 Correct answer is option 'B'. Can you explain this answer?.
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