Vieta's Formula

Vieta's Formulas provide a way to relate the coefficients of a polynomial to the sums and products of its roots. Named after the French mathematician François Viète, these formulas establish a connection between the roots of a polynomial and its coefficients.

If you have a polynomial equation and know its roots, Vieta's formulas help you determine the relationships between the equation's coefficients and its roots.

For example: Consider the equation f(x) = x² − x − 6, which factors to (x − 3) (x + 2), so the roots are x = 3 and x = -2.

• Vieta's formula can find that the sum of the roots is −b/a = 1 and the product is c/a = −6, which matches the actual values: 3 + (−2) = 1, and 3⋅(−2) = −6.
• Vieta’s formulas make it easy to find these relationships directly from the equation without factoring.

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, the quadratic equation is given by :

• x² - (sum of roots)x + (product of roots) = 0

Example: Solve for the roots using Vieta's formulas for the equation: x² − 5x + 6 = 0.
Solution:

From the equation, we can identify a = 1, b = −5b , and c = 6.
Sum of roots = α + β = -b/a = 5
Product of roots = αβ = c/a = 6

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

Example: Solve for the roots using Vieta's formulas for the equation: x³ − 6x² + 11x − 6 = 0.
Solution:

From the equation, we can identify a=1, b=−6, c=11, and d=−6.
Sum of roots = α + β + γ = -b/a = 6
Sum of product of two roots = αβ + αγ + βγ = c/a = 11
Product of roots = αβγ = -d/a = 6

Vieta's Formula for Higher Degree Polynomials

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,
The sum of the roots: r₁ + r₂ + r₃ +.......... + r_n-1 + r_n = -a_n-1/a_n
The sum of the products of the roots taken two at a time: (r₁r₂ + r₁r₃ +.... +r₁r_n) + (r₂r₃ + r₂r₄ +....... +r₂r_n) + ......... + r_n-1r_n = a_n-2/a_n
:
:
The product of all the roots: r₁r₂...r_n = (-1)ⁿ(a₀/a_n) 

Example: Let's take an example of a degree n = 5 polynomial and apply Vieta’s formulas.

Solution:

This is a degree-5 polynomial with leading coefficient a₅ = 1, and the coefficients are:

r₁ + r₂ + r₃ + r₄ + r₅ = -a₄/a₅ = -(4) = 4
r₁ r₂ + r₁ r₃ +...+ r₄ r₅ = a₃/a₅ = 3
r₁ r₂ r₃ ...... = -a₂/a₅ = -2
r₁ r₂ r₃ r₄...... = a₁/a₅ = -1
r₁ r₂ r₃ r₄ r₅ = (-1)⁵a₀/a₅ = -6

Solved Questions on Vieta's Formula

Question 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

Question 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

Question 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

Question 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

Question 5: If α, β and γ are the roots of the equation and given that α + β +  γ= 10, αβ + αγ + βγ = -1, and αβγ = -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

Question 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

Question 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