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
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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
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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
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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
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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) = anxn+ an-1xn-1 + an-2xn-2 + ……… + a2x2 + a1x +a0 is a quadratic equation with roots r1, r2, r3, …… rn-1, rn then,
r1 + r2 + r3 +………. + rn-1 + rn = -an-1/an
(r1r2 + r1r3 +…. +r1rn) + (r2r3 + r2r4 +……. +r2rn) + ……… + rn-1rn = an-2/an
:
:
r1r2…rn = (-1)n(a0/an)
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