Content Curator
Vieta’s formulas were discovered by Francois Viète, a French mathematician whose work on new algebra was an important step towards modern algebra. Vieta’s formulas are formulas that relate the coefficients of a polynomial to the sums and products of its roots. Their simplest applications are in quadratics and algebra.
Table of Contents
| Table of Content |
Key Terms: Vieta’s Formula, Quadratic Equations, Polynomials, Algebra, Francois Viète, Quadratics, Coefficients, Roots.
Vieta’s Formula for Quadratics
[Click Here for Sample Questions]
Consider the quadratic equation f(x) = ax2+ bx + c with roots r1 and r2 when f(x) = 0. Then, r1+ r2 = \(\frac{-b}{a}\) and r1 r2 = \(\frac{c}{a}\).
Proof: Let p and q be the real roots of the monic quadratic equation x2+ bx+c = 0. A monic quadratic equation is used because it is easy to divide the whole equation by its leading coefficient to get a monic version of it.
Since p and q are roots of the equation,
x2+bx+c ≡ (x-p)(x-q)
x2+bx+c ≡ x2 -(p+q)x +pq
Since two polynomials are equal if and only if their coefficients are equal b= -(p+q) and c=pq.
The roots can be generalized to include complex numbers. i.e., Given two complex numbers p and q, a mono quadratic x2-(p+q)x+pq=0, with roots p and q can be constructed.
To find out when the coefficients of this mono quadratic will be real, set p= p1+p2i and q=q1+q2i.
Then, the coefficients are b=p1+ q1 -(p2+ q2)i and c=p1q1- p2q2 + (p1q2-p2q1)i.
For b to be real, p2+ q2 =0 ⇒ p2= -q2.
For c to be real, p1q2+ p2q1=0 ⇒ q2p1- q1=0. This implies that either q2=0 or p1= q1. Hence, either p and q are real or they are complex conjugates of each other.
The video below explains this:
Polynomials Detailed Video Explanation:
Read More:
Vieta’s Formula for Generalized Higher Degree Polynomials
[Click Here for Sample Questions]
Consider a polynomial of degree n, P(x) = anxn+ an-1xn-1+ ... +a1x+a0, with complex coefficients and having complex roots, r1, r2, …, rn-1, rn. The (n-k)th coefficient an-k is related to a signed sum of all possible sub products of roots, taken k at-a-time as follows.
\(\displaystyle\sum_{i_1<i_2<....<i_k{\leq}n}^{} r_{i_1} r_{i_2}... r_{i_k}\)= (-1)k\(\frac{a_{n-k}}{a_k}\) for k = 1, 2, 3, …, n.
Proof: Consider a polynomial, P(x) = anxn+ an-1xn-1+ ... +a1x+a0, of degree n that has complex roots r1, r2, …, rn-1, rn. The expressions on the left side of Vieta’s formula are the elementary symmetric functions of r1, r2, …, rn-1, rn. Comparing the coefficients in the equation, anxn+ an-1xn-1+ ... +a1x+a0= an(x-r1)(x-r2)...(x- rn).
Hence, Vieta’s formula gives the equation to find the sum of the roots.
\(\displaystyle\sum_{i=r}^{n} r_i\)= - \(\frac{a_{n-1}}{a_n}\)
Also, the following equation can find the products of the roots.
r1, r2, …, rn-1, rn= (-1)n\(\frac{a^o}{a^n}\)
Things to Remember
- Vieta’s formulas were discovered by Francois Viète, a French mathematician whose work on new algebra was an important step towards modern algebra.
- The simplest applications of Vieta’s formulas are quadratics and algebra.
- Vieta’s formulas are formulas that relate the coefficients of a polynomial to the sums and products of its roots.
- Consider the quadratic equation f(x) = ax2+ bx + c with roots r1 and r2 when f(x) = 0. Then, r1+ r2 = \(\frac{-b}{a}\) and r1 r2 = \(\frac{c}{a}\).
- Consider a polynomial of degree n, P(x) = anxn+ an-1xn-1+ ... +a1x+a0, with complex coefficients and having complex roots r1, r2, …, rn-1, rn. The (n-k)th coefficient an-k is related to a signed sum of all possible sub products of roots, taken k at-a-time as follows:
\(\displaystyle\sum_{i_1<i_2<....<i_k{\leq}n}^{} r_{i_1} r_{i_2}... r_{i_k}\)= (-1)k\(\frac{a_{n-k}}{a_k}\) for k = 1, 2, 3, …, n.
Sample Questions
Ques. Determine the sum and product of the polynomial x2-11x+22 using Vieta’s formula. (3 Marks)
Ans. Using Vieta’s formula,
Sum of the roots = - \(\frac{\text{-Coefficient of x}}{\text{Coefficient of x^2}}\) = \(\frac{-(-11)}{1}\) = 11
Product of the roots = \(\frac{\text{constant}}{\text{Coefficient of x^2}}\)=\(\frac{22}{1}\)= 22
Ques. Determine the polynomial P(x) if the sum and product of its coefficients are 9 respectively and P(6) = 4. (3 Marks)
Ans. Using Vieta’s formula, we can write the polynomial as follows.
P(x): k(x2-9x+20)
Since P(6) = 2, P(6) = 4 k(62 - 9(6) + 20) = 4 k(36 - 54 + 20) = 4 2k = 4 k = 2
Therefore, the polynomial is P(x): 2x2 - 18x + 40.
Ques. Find the value of \(\displaystyle\sum_{}^{} \)\(\frac{1}{\beta\gamma}\) in terms of coefficients if α, β and are the roots of the equation x3+px2+qx+r=0. (3 Marks)
Ans. Given that α, β and γ are the roots of the equation,
Ques. Find the equation that has the same roots as the cubic equation x3 + ax2+ bx + c = 0 (5 Marks)
Ans. Let α, β, and γ be the roots of x3 + ax2 + bx + c = 0.
Then, we get
\(\displaystyle\sum_{}^{} 1\) = α + β + γ = -a, …. (1)
\(\displaystyle\sum_{}^{} 2\) = αβ + βγ + γα = b, …. (2)
\(\displaystyle\sum_{}^{} 3\) = αβγ = -c. …. (3)
We are looking for an equation whose roots are α2, β2, and γ2.
Using (1), (2), and (3),
Hence, the required equation is
x3 – (α 2 + β2 + γ2) x2 + (α 2 β2 + β2γ2 + γ2α2) x - α2β2γ2 = 0.
That is, x3 - (a2 - 2b) x2 + (b2 - 2ca) x - c2 = 0.
Ques. Determine the value of \(\frac{1}{x_1}\)+ \(\frac{1}{x_2}\) if x1 and x2 are roots of the equation x2 +9x+33. (3 Marks)
Ans. Using Vieta’s formulas, x1+x2=-9 and x1x2 = 33.
Hence, \(\frac{1}{x_1}\)+ \(\frac{1}{x_2}\) =\(\frac{x_1+x_2}{x_1x_2}\) = \(\frac{-9}{33}\)
Ques. State and prove Vieta’s formula for quadratic equations. (4 Marks)
Ans. For the quadratic equation f(x) = ax2+ bx + c with roots r1 and r2 when f(x) = 0, r1+r2 = \(\frac{-b}{a}\)and r1r2 = \(\frac{c}{a}\).
Proof
Let p and q be the real roots of the monic quadratic equation x2+ bx + c = 0. A monic quadratic equation is used because it is easy to divide the whole equation by its leading coefficient to get a monic version of it.
Since p and q are roots of the equation,
x2+ bx + c ≡ (x-p)(x-q)
x2+ bx + c ≡ (p+q)x+pq
Since two polynomials are equal if and only if their coefficients are equal b= -(p+q) and c=pq.
Ques. How can Vieta’s formula for quadratic equations be extended to complex numbers? (4 Marks)
Ans. Vieta’s formula can be generalized to include complex numbers. i.e., Given two complex numbers p and q, a mono quadratic x2-(p+q)x+pq=0, with roots p and q can be constructed.
To find out when the coefficients of this mono quadratic will be real, set p= p1+p2i and q=q1+q2i.
Then, the coefficients are
b=p1+ q1-(p2+ q2)i and c=p1q1- p2q2+(p1q2-p2q1)i.
For b to be real, p2+ q2 =0 ⇒ p2= -q2.
For c to be real, p1q2+p2q1=0 ⇒ q2p1- q1=0. This implies that either q2=0 or p1= q1. Hence, either p and q are real or they are complex conjugates of each other.
Ques. Find the sum of the roots and the product of the roots of the polynomial P(x) = 3x3 +2x2 -3x+ 5. (2 Marks)
Ans. Using Vieta’s formula, r1+r2 + r3 = \(\frac{-a_2}{a_1}\)= \(\frac{-2}{3}\) and r1r2 = \(\frac{a_o}{a_3}\)= \(\frac{5}{3}\)
Ques. Prove Vieta’s formula for generalized higher degree polynomials. (4 Marks)
Ans. Consider a polynomial, P(x) = anxn+ an-1xn-1+ ... +a1x+a0, of degree n that has complex roots r1, r2, …, rn-1, rn. The expressions on the left side of Vieta’s formula are the elementary symmetric functions of r1, r2, …, rn-1, rn. Comparing the coefficients in the equation, anxn+ an-1xn-1+ ... +a1x+a0= an(x-r1)(x-r2)...(x- rn).
Hence, Vieta’s formula gives the equation to find the sum of the roots.
\(\displaystyle\sum_{i=r}^{n} r_i\)= - \(\frac{a_{n-1}}{a_n}\)
Also, the following equation can find the products of the roots.
r1, r2, …, rn-1, rn= (-1)n\(\frac{a^o}{a^n}\)
Also read:
Comments