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Last updated on October 4, 2025
Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. These formulas are named after the French mathematician François Viète. In this topic, we will learn about Vieta's formulas and how they can be applied to polynomials.
Vieta's formulas provide a powerful tool for relating the coefficients of a polynomial to the sums and products of its roots. Let’s explore how to apply Vieta's formulas to find these relationships.
For a quadratic equation of the form\( ( ax^2 + bx + c = 0 ),\) Vieta's formulas are: - The sum of the roots \(r_1 \) and \( r_2 \) is given by: \(( r_1 + r_2 = -\frac{b}{a} )\)
The product of the roots is given by:\( ( r_1 \cdot r_2 = \frac{c}{a} )\)
For a cubic equation of the form \(( ax^3 + bx^2 + cx + d = 0 )\), Vieta's formulas are:
The sum of the roots\( ( r_1, r_2, r_3 )\) is: (\(( r_1 + r_2 + r_3 = -\frac{b}{a} )
\)
The sum of the products of the roots taken two at a time is: \(( r_1r_2 + r_2r_3 + r_1r_3 = \frac{c}{a} ) \)
The product of the roots is: \(( r_1 \cdot r_2 \cdot r_3 = -\frac{d}{a} )\)
For a polynomial of degree n :
The sum of the roots taken one at a time is: \(( -\frac{\text{coefficient of } x^{n-1}}{\text{leading coefficient}} )\)
The sum of the products of the roots taken two at a time is: \(( \frac{\text{coefficient of } x^{n-2}}{\text{leading coefficient}} ) \) And so on, until the product of the roots (with alternating signs) is given by the constant term divided by the leading coefficient.
Vieta's formulas are crucial in algebra and allow for the analysis and understanding of polynomial roots without explicitly solving the polynomial.
They provide insights into relationships between coefficients and roots, useful in various mathematical and real-world applications.
When applying Vieta's formulas, students might encounter several pitfalls. Here are some mistakes and ways to avoid them:
Find the sum and product of the roots of the equation \( 3x^2 - 5x + 2 = 0 \).
The sum of the roots is\( (\frac{5}{3}) \)and the product is \((\frac{2}{3}).\)
For the equation \(( 3x^2 - 5x + 2 = 0 ):\)
The sum of the roots: \(( r_1 + r_2 = -\frac{-5}{3} = \frac{5}{3} )\)
The product of the roots:\( ( r_1 \cdot r_2 = \frac{2}{3} )\)
For the cubic polynomial \( x^3 - 6x^2 + 11x - 6 = 0 \), use Vieta's formulas to find the sum of the roots.
The sum of the roots is 6.
The sum of the roots of the cubic polynomial \(( x^3 - 6x^2 + 11x - 6 = 0 ) \) is given by:\( ( r_1 + r_2 + r_3 = -\frac{-6}{1} = 6 )\)
Determine the product of the roots for the polynomial ( 2x³- 3x² + x - 5 = 0 ).
The product of the roots is 5/2.
The product of the roots for \(( 2x^3 - 3x^2 + x - 5 = 0 ) \) is: \(( r_1 \cdot r_2 \cdot r_3 = -\frac{-5}{2} = \frac{5}{2} )\)
Given the quadratic equation 4x² + 8x + 3 = 0 , find the sum and product of the roots.
The sum of the roots is -2 and the product is 3/4.
For the equation\( ( 4x^2 + 8x + 3 = 0 ):\) - The sum of the roots: \(( r_1 + r_2 = -\frac{8}{4} = -2 ) \)
The product of the roots: \(( r_1 \cdot r_2 = \frac{3}{4} )\)
For the equation ( x³+ 7x² + 14x + 8 = 0 ), use Vieta's formulas to determine the sum of the roots.
The sum of the roots is -7.
The sum of the roots for the cubic equation\( x^3 + 7x^2 + 14x + 8 = 0 \) is: \(( r_1 + r_2 + r_3 = -\frac{7}{1} = -7 )\)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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