Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > The equation 7x7+ 14x6+ 12 x5+ 3x4+ 12x3+ 10x...
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The equation 7x7 + 14x6 + 12 x5 + 3x4 + 12x3+ 10x2 + 5x+ 7 = 0 has
- a)All complex roots
- b)At least one real root
- c)Four pairs of imaginary roots
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The equation 7x7+ 14x6+ 12 x5+ 3x4+ 12x3+ 10x2+ 5x+ 7 = 0hasa)All comp...
Since the polynomial has highest degree 7. So there are 7 roots possible for it
now suppose if an imaginary number a+bi is also root of this polynomial then a-bi will also be the root of this polynomial
That means there must be even number of complex root possible becoz they occur in pair.
Now we will solve this question option wise
A) All complex root
This is not possible. The polynomial has 7 roots and as I mention a polynomial should have even number of complex root and 7 is not even. So this option is wrong
B) At least one real root
This is possible. Since polynomial has 7 roots and only even number of complex root is possible, that means this polynomial has max 6 complex roots and Hence minimum one real root. So this option is correct
C) Four pairs of imaginary roots 4 pair means 8 complex root. But this polynomial can have atmost 7 roots. So this option is also wrong
now suppose if an imaginary number a+bi is also root of this polynomial then a-bi will also be the root of this polynomial
That means there must be even number of complex root possible becoz they occur in pair.
Now we will solve this question option wise
A) All complex root
This is not possible. The polynomial has 7 roots and as I mention a polynomial should have even number of complex root and 7 is not even. So this option is wrong
B) At least one real root
This is possible. Since polynomial has 7 roots and only even number of complex root is possible, that means this polynomial has max 6 complex roots and Hence minimum one real root. So this option is correct
C) Four pairs of imaginary roots 4 pair means 8 complex root. But this polynomial can have atmost 7 roots. So this option is also wrong
Most Upvoted Answer
The equation 7x7+ 14x6+ 12 x5+ 3x4+ 12x3+ 10x2+ 5x+ 7 = 0hasa)All comp...
Understanding the Polynomial Equation
The equation provided is a seventh-degree polynomial:
7x^7 + 14x^6 + 12x^5 + 3x^4 + 12x^3 + 10x^2 + 5x + 7 = 0.
Analyzing the roots of this polynomial is essential to determine the nature of its solutions.
Nature of Roots
- Degree of the Polynomial: The polynomial is of degree 7, which means it can have up to 7 roots, including real and complex roots.
- Real Roots vs. Complex Roots: A polynomial with real coefficients must have either real roots or complex conjugate pairs. This implies that if there is at least one real root, the total number of roots can include complex roots.
Applying Descartes' Rule of Signs
- Positive Roots: Analyzing the sign changes for positive x values, we find several sign changes, indicating potential positive real roots.
- Negative Roots: For negative x values, we observe fewer sign changes, suggesting fewer negative real roots.
Conclusion on Roots
- Given that there are indications of sign changes, it is likely that the polynomial has at least one real root.
- Therefore, option B (“At least one real root”) is correct, as a polynomial of degree 7 is guaranteed to have at least one real root, owing to the fundamental theorem of algebra.
Final Note
- Since the polynomial has real coefficients, any complex roots would indeed occur in conjugate pairs. Thus, confirming option B is the most accurate choice regarding the nature of the roots.
The equation provided is a seventh-degree polynomial:
7x^7 + 14x^6 + 12x^5 + 3x^4 + 12x^3 + 10x^2 + 5x + 7 = 0.
Analyzing the roots of this polynomial is essential to determine the nature of its solutions.
Nature of Roots
- Degree of the Polynomial: The polynomial is of degree 7, which means it can have up to 7 roots, including real and complex roots.
- Real Roots vs. Complex Roots: A polynomial with real coefficients must have either real roots or complex conjugate pairs. This implies that if there is at least one real root, the total number of roots can include complex roots.
Applying Descartes' Rule of Signs
- Positive Roots: Analyzing the sign changes for positive x values, we find several sign changes, indicating potential positive real roots.
- Negative Roots: For negative x values, we observe fewer sign changes, suggesting fewer negative real roots.
Conclusion on Roots
- Given that there are indications of sign changes, it is likely that the polynomial has at least one real root.
- Therefore, option B (“At least one real root”) is correct, as a polynomial of degree 7 is guaranteed to have at least one real root, owing to the fundamental theorem of algebra.
Final Note
- Since the polynomial has real coefficients, any complex roots would indeed occur in conjugate pairs. Thus, confirming option B is the most accurate choice regarding the nature of the roots.
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The equation 7x7+ 14x6+ 12 x5+ 3x4+ 12x3+ 10x2+ 5x+ 7 = 0hasa)All complex rootsb)At least one real rootc)Four pairs of imaginary rootsd)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
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