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The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative value of r?
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The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the r...
Given Equation
The polynomial is given as:
\[ x^3 + (2 + 1)x^2 + (4r - 1)x + 2 = 0 \]
This can be simplified to:
\[ x^3 + 3x^2 + (4r - 1)x + 2 = 0 \]
Substituting the Root
Since -2 is a root, substituting \( x = -2 \) into the equation yields:
\[ (-2)^3 + 3(-2)^2 + (4r - 1)(-2) + 2 = 0 \]
\[ -8 + 3(4) - 2(4r - 1) + 2 = 0 \]
\[ -8 + 12 - 8r + 2 = 0 \]
Simplifying gives:
\[ 6 - 8r = 0 \]
Thus,
\[ r = \frac{6}{8} = \frac{3}{4} \]
Finding Other Roots
To check the nature of the other roots, we can use the condition for real roots. For the cubic polynomial to have three real roots, the discriminant must be non-negative.
Formulating the Discriminant
The polynomial can be analyzed via its derivative:
\[ p'(x) = 3x^2 + 6x + (4r - 1) \]
Setting \( p'(-2) \):
\[ 3(-2)^2 + 6(-2) + (4r - 1) = 0 \]
\[ 12 - 12 + (4r - 1) = 0 \]
This implies:
\[ 4r - 1 = 0 \rightarrow r = \frac{1}{4} \]
Conclusion
To ensure all roots are real and one is -2, the minimum non-negative value of \( r \) that satisfies both conditions is:
\[ r = \frac{3}{4} \]
Thus, the minimum possible non-negative value of \( r \) is \( \frac{3}{4} \).
The polynomial is given as:
\[ x^3 + (2 + 1)x^2 + (4r - 1)x + 2 = 0 \]
This can be simplified to:
\[ x^3 + 3x^2 + (4r - 1)x + 2 = 0 \]
Substituting the Root
Since -2 is a root, substituting \( x = -2 \) into the equation yields:
\[ (-2)^3 + 3(-2)^2 + (4r - 1)(-2) + 2 = 0 \]
\[ -8 + 3(4) - 2(4r - 1) + 2 = 0 \]
\[ -8 + 12 - 8r + 2 = 0 \]
Simplifying gives:
\[ 6 - 8r = 0 \]
Thus,
\[ r = \frac{6}{8} = \frac{3}{4} \]
Finding Other Roots
To check the nature of the other roots, we can use the condition for real roots. For the cubic polynomial to have three real roots, the discriminant must be non-negative.
Formulating the Discriminant
The polynomial can be analyzed via its derivative:
\[ p'(x) = 3x^2 + 6x + (4r - 1) \]
Setting \( p'(-2) \):
\[ 3(-2)^2 + 6(-2) + (4r - 1) = 0 \]
\[ 12 - 12 + (4r - 1) = 0 \]
This implies:
\[ 4r - 1 = 0 \rightarrow r = \frac{1}{4} \]
Conclusion
To ensure all roots are real and one is -2, the minimum non-negative value of \( r \) that satisfies both conditions is:
\[ r = \frac{3}{4} \]
Thus, the minimum possible non-negative value of \( r \) is \( \frac{3}{4} \).
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The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative value of r?
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The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative value of r? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative value of r? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation x³ + (2 + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative value of r?.
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