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find all real numbers a such that the roots of the polynomial x³ - 3x² + 7x + a form an arithmetic progression and are not all real.

User Jmlane
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2 Answers

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Final answer:

To find all real numbers a for the given polynomial where the roots are in an arithmetic progression, we use Vieta's formulas assuming the roots are r-d, r, and r+d with r = 1. Hence, a is any real number representable as 1 - d², where d is a real number.

Step-by-step explanation:

To find all real numbers a such that the roots of the polynomial x³ - 3x² + 7x + a form an arithmetic progression and are not all real, we must consider the properties of roots in an arithmetic progression and the complex roots theorem. A polynomial of degree 3 always has three roots, counting multiplicities. In any arithmetic sequence, if the common difference is d, the three terms can be represented as r-d, r, and r+d, where r is the middle term.



Since one root must be complex, and complex roots of polynomials with real coefficients come in conjugate pairs, there can't be exactly one complex root. Consequently, all three roots must be real. However, the question stipulates that not all roots are real; this implies an inconsistency or typo in the question, and the condition that the roots are not all real should be ignored or clarified.



For roots that are in arithmetic progression, let the roots be r-d, r, and r+d. Now, by Vieta's formulas:





The constant term a of the polynomial can be found by the product of its roots:




Therefore, the value of a is any number that can be represented in the form 1 - d², where d is a real number.

User Tropicalrambler
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For every real number
\( a \) less than or equal to
\( -1 \), the polynomial will have one real root (which is
\( 1 \)) and two non-real roots that are complex conjugates of each other, and these roots will be in arithmetic progression.

To find the real numbers
\( a \) such that the roots of the polynomial
\( x^3 - 3x^2 + 7x + a \) form an arithmetic progression and are not all real, we'll need to determine the conditions that these roots must satisfy.

Let's say the three roots are
\( r \), \( r+d \), and
\( r-d \), where
\( r \) is the middle root of the arithmetic progression and
\( d \) is the common difference. Since not all roots are real, one root must be real (the middle root
\( r \)) and the other two must be complex conjugates.

The sum of the roots, given by Vieta's formulas, is equal to the coefficient of
\( x^2 \) with the opposite sign. For this polynomial, the sum of the roots is . So we have:


\[ r + (r+d) + (r-d) = 3 \]


\[ 3r = 3 \]


\[ r = 1 \]

The middle root
\( r \) is 1. So the roots are
\( 1 \),
\( 1+d \), and
\( 1-d \). Because two of the roots are complex conjugates,
\( d \) must be imaginary. Let
\( d = bi \), where
\( b \) is a real number and
\( i \) is the imaginary unit.

Now we consider the product of the roots. For complex conjugates,
\( (1+bi)(1-bi) = 1 - b^2i^2 = 1+b^2 \), because
\( i^2 = -1 \). The product of the roots is given by the constant term
\( a \) of the polynomial with the opposite sign:


\[ (1)(1+bi)(1-bi) = -a \]


\[ 1+b^2 = -a \]


\[ a = -1 - b^2 \]

The polynomial is
\( x^3 - 3x^2 + 7x + a \), and we want to ensure that when
\( a = -1 - b^2 \), the polynomial has one real root and two complex conjugate roots. However, since we already established that the roots are in arithmetic progression with a common difference of \( bi \), we have effectively ensured that the non-real roots are complex conjugates by construction.

So the possible values for
\( a \) are those for which
\( b \) is a real number. That is,
\( a \) can take any value of the form
\( -1 - b^2 \) where
\( b \) is a real number. This gives us a range of values for
\( a \), since
\( b^2 \) is always non-negative:


\[ a \leq -1 \]

For every real number
\( a \) less than or equal to
\( -1 \), the polynomial will have one real root (which is
\( 1 \)) and two non-real roots that are complex conjugates of each other, and these roots will be in arithmetic progression.

User Nikhil J Joshi
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