Vieta’s Formula

Algebra is one of the basic topics of mathematics. Polynomials are an essential part of algebra. Vieta’s formula is used in polynomials. This article is about Vieta’s formula which relates the sum and product of roots to the coefficient of the polynomial. This formula is specifically used in algebra.

Vieta’s Formula

Vieta’s formulas are those formulas that provide the relation between the sum and product of roots of the polynomial with the coefficients of the polynomials. Vieta’s formula describes the coefficients of the polynomial in the form of the sum and product of its root.

Vieta’s Formula

Vieta’s formula deals with the sum and product of the roots and the coefficient of the polynomial. It is used when we have to find the polynomial when roots are given or we have to find the sum or product of the roots.

Vieta’s Formula for Quadratic Equation

• If f(x) = ax² + bx + c is a quadratic equation with roots α and β then,
  • Sum of roots = α + β = -b/a
  • Product of roots = αβ = c/a
• If the sum and product of roots are given then, the quadratic equation is given by :
  • x² – (sum of roots)x + (product of roots) = 0

Vieta’s Formula for the Cubic Equation

• If f(x) = ax³+ bx² + cx +d is a quadratic equation with roots α, β and γ then,
  • Sum of roots = α + β + γ = -b/a
  • Sum of product of two roots = αβ + αγ + βγ = c/a 
  • Product of roots = αβγ = -d/a
• If the sum and product of roots are given then, the cubic equation is given by :
  • x³ – (sum of roots)x² + (sum of product of two roots)x – (product of roots) = 0

Vieta’s Formula for Generalized Equation

If f(x) = a_nxⁿ+ a_n-1x^n-1 + a_n-2x^n-2 + ……… + a₂x² + a₁x +a₀ is a quadratic equation with roots r₁, r₂, r₃, …… r_n-1, r_n then,

r₁ + r₂ + r₃ +………. + r_n-1 + r_n = -a_n-1/a_n

(r₁r₂ + r₁r₃ +…. +r₁r_n) + (r₂r₃ + r₂r₄ +……. +r₂r_n) + ……… + r_n-1r_n = a_n-2/a_n

:

:

r₁r₂…r_n = (-1)ⁿ(a₀/a_n) 

Sample Problems

Problem 1: If α , β are the roots of the equation : x² – 10x + 5 = 0 , then find the value of (α² + β²)/(α²β  + αβ²).

Solution: 

Given Equation:

• x² – 10x + 5 = 0

By Vieta’s Formula

α + β = -b/a = -(-10)/1 = 10

αβ = c/a = 5/1 = 5

As (α²+β²) = (α + β )² – 2αβ

= (10)² – 2×5 

= 100 – 10  

(α²+β²)  = 90

Now value of (α² + β²)/(α²β  + αβ²)

= (α² + β²)/(αβ(α + β))

= 90/(5×10)

= 90/50

= 1.8

Problem 2: If α , β are the roots of the equation : x² + 7x + 2 = 0 , then find the value of 14÷(1/α + 1/ β).

Solution:

Given Equation:

• x² + 7x + 2 = 0

By Vieta’s Formula

α + β = -b/a = -7/1 = -7

αβ = c/a = 2/1 = 2

Now, (1/α + 1/ β) = (α + β)/αβ

(1/α + 1/ β) = -7/2

Now value of  14÷(1/α + 1/ β)

= 14 ÷ (-7/2)

= 14 × (-2/7)

= -4

Problem 3: If α , β are the roots of the equation : x² + 10x + 2 = 0 , then find the value of (α/β +  β/α).

Solution: 

Given Equation:

• x² + 10x + 2 = 0

By Vieta’s Formula

α + β = -b/a = 10/1 = 10

αβ = c/a = 2/1 = 2

As (α²+β²) = (α + β )² – 2αβ

= 10² – 2×2

= 100 – 4 

= 96

Now value of (α/β +  β/α) = (α²+β²)/αβ 

= 96/2

= 48

Problem 4: If α and β are the roots of the equation and given that α + β = -100 and αβ = -20 then find the quadratic equation.

Solution:

Given, 

• Sum of roots = α + β = -100
• Product of roots = αβ = -20

Quadratic equation is given by:

x² – (sum of roots)x + (product of roots) = 0

x² – (-100)x + (-20) = 0

x² + 100x – 20 = 0

Problem 5: If α , β and  γ are the roots of the equation and given that α + β +  γ= 10, αβ + αγ + βγ = -1 and  αβ γ = -6 then find the cubic equation.

Solution:

Given,

• Sum of roots = α + β + γ= 10,
• Sum of product of two roots = αβ + αγ + βγ = -1
• Product of roots = αβγ = -6

Cubic equation is given by:

x³ – (sum of roots)x² + (sum of product of two roots)x – (product of roots) = 0

x³ – 10x² + (-1)x – (-6) = 0

x³ – 10x² – x + 6 = 0

Problem 6: If α , β and  γ are the roots of the equation x³ + 1569x² – 3 = 0 then find the value of [(1/α) + (1/β )]³ + [(1/γ) + (1/β )]³ + [(1/γ) + (1/α )]³

Solution:

Given,

• Sum of roots = α + β + γ= -b/a = -1569/1 = -1569
• Sum of product of two roots = αβ + αγ + βγ = c/a = 0/1 = 0
• Product of roots = αβγ = -d/a = -(-3)/1 = 3

Since, (p³ + q³ + r³ – 3pqr) = (p + q + r)(p² +q² + r² – pq – qr – pr)                 ……(1)

Let, p = (1/α) + (1/β )  , q = (1/γ) + (1/β ) , r = (1/γ) + (1/α )

p + q + r = 2[(1/α) + (1/β ) +  (1/γ) ] = 2(αβ + αγ + βγ)/αβγ

= 2(0/3) = 0 

From equation (1):

(p³ + q³ + r³ – 3pqr) = 0

p³ + q³ + r³ = 3pqr 

[(1/α) + (1/β )]³ + [(1/γ) + (1/β )]³ + [(1/γ) + (1/α )]³ = 3[(1/α) + (1/β )][(1/γ) + (1/β )][(1/γ) + (1/α )]

= 3(-1/γ)(-1/α) (-1/β )

= -3/αβγ  = -3/3

= -1

Problem 7: If α and β are the roots of the equation x² – 3x +2 =0 then find the value of α² – β².

Solution:

Given,

• Sum of roots = α + β = -b/a = -(-3)/1 = 3
• Product of roots = αβγ = c/a = 2/1 = 2

As (α – β)² = (α + β)² -4αβ  

(α – β)² = (3)² – 4(2) = 9 – 8 = 1

(α – β) = 1

Since, 

α² – β² = (α – β)(α + β) = (1)(3) = 3

α² – β² = 3