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Vieta's Formula Vieta's formula relates the coecients of polynomials to the sums and products of their roots, as well as the products of the roots taken in groups.

f (x) = x2 + 2x − 15, it will have roots of x = −5and x = 3, because f (x) = x + 2x − 15 = (x − 3)(x + 5). Vieta's formula can find the sum of the roots (3 + (−5) = −2) and the product of the roots (3 ⋅ (−5) = −15) without finding each root directly. While this is fairly trivial in this specific example, Vieta's formula For example, if there is a quadratic polynomial 2

is extremely useful in more complicated algebraic polynomials with many roots or when the roots of a polynomial are not easy to derive. For some problems, Vieta's formula can serve as a shortcut to finding solutions quickly knowing the sums or products of their roots.

Contents Vieta's Formula - Quadratic Equations Vieta's Formula - Forming Quadratics Generalization to Higher Degree Polynomials Vieta's Formula Problem Solving - Basic Vieta's Formula Problem Solving - Intermediate Vieta's Formula Problem Solving - Advanced Vieta Root Jumping See Also

Vieta's Formula - Quadratic Equations Let's start with a definition. D EFI NITI ON

Vieta's Formula for Quadratics: Given

f (x) = ax2 + bx + c, if the equation f (x) = 0 has roots r1 and r 2, then b r1 + r2 = − , a

r1 r2 =

c . a □

The proof of this statement is given at the end of this section. We immediately see how the coecients of a quadratic help to determine the relationship of the roots. EXAMPLE

If α and β are the roots of the quadratic .

α+β

.

αβ

.

α2 + β2 ?

x2 − 4x + 9 = 0, what are the values of

. From Vieta's formula, we recognize that

α + β = 4.

. From Vieta's formula, we recognize that

αβ = 9.

. Vieta's formula doesn't tell us the value of

α2 + β 2 directly. What we need to do, is to write α2 + β 2in terms of α + β

and/or αβ , and we can then substitute these values in. We have

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α + β2

= (α + β)2 − 2αβ = 42 − 2 × 9 = −2.□

Note: In this question, the roots are the complex numbers

2±

5 i. If we were to use these as the values of αand β and

then compute, we have a higher chance of making careless calculation mistakes. Vieta's formula oers us a simpler approach.

EXAMPLE

Can you quickly guess the roots of the quadratic

x 2 − 5x + 6?

If p and q are the roots of the given equation, then Vieta's formula tells us

p + q = 5, and it's not hard to see that 2 + 3

pq = 6,

= 5 and 2 × 3 = 6. So the roots must be 2 and 3, and they indeed are. □

EXAMPLE

Given

α and β are the roots of the quadratic ax 2 + bx + c = 0, express −b

Vieta's formula gives us a

= α + β and

c a

= αβ. Substituting these in, we get b 2−4ac a2

You may recognize b 2

b2 − 4ac in terms of α and β . a2

2

= ba2 − 4 ca = (α + β)2 − 4αβ = α 2 + 2αβ + β 2 − 4αβ = α 2 − 2αβ + β 2 = (α − β)2 .□

− 4ac from the Quadratic Formula. In fact, we can show that −b ±

1 −b b2 − 4ac = ( ± a 2a 2

b2 − 4ac (α + β) ± ∣α − β∣ )= = αorβ , a2 2

which proves the quadratic formula. PROOF

By the Remainder Factor Theorem, since the polynomial f (x) has roots r 1 and r2, it must have the form f (x) = A(x − r 1)(x − r2 ) = Ax 2 − A(r1 + r2)x + Ar1 r2 for some constant A. Comparing coecients with f (x)

= ax2 + bx + c, we conclude that a = A, b = −A(r1 + r2 ) and c = Ar1 r2. Hence,

we get

b r1 + r2 = − , a

r1 r2 =

c .□ a

Vieta's Formula - Forming Quadratics Let p and q be the real roots of the monic quadratic equation

x2 + bx + c = 0, (b, c) ∈ R 2. Why monic? Because we can always divide the whole equation by its leading coecient to get a monic version of it. So what can we say about

b, c? Since p and q are the roots of this equation, we can factorize the equation as x 2 + bx + c ≡ (x − p) (x − q ) .

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Expanding the right side and rearranging, we find

x2 + bx + c ≡ x 2 − (p + q )x + pq . Since two polynomials are equal if and only if their coecients are equal, by equating the coecients we get

b = −(p + q),

c = pq.

This is the so-called Vieta's formula for a quadratic polynomial. It can be similarly extended to polynomials of higher degree. The roots can be generalized to include complex numbers. That is, given two complex numbers a monic quadratic whose roots are

p and q, we can always construct

p and q. More specifically, the quadratic will be x2 − (p + q )x + pq = 0.

When will both the coecients be real? To find the answer, set

b = − (p 1 + q 1 ) − (p 2 + q 2 ) i, For b to be real, we need p 2

p = p 1 + p2 iand q = q1 + q2 i. Then the coecients are c = (p 1q 1 − p 2q 2 ) + (p 1q 2 + p 2q 1 )i.

+ q2 = 0 ⇒ p2 = −q 2. For c to be real, we need p1 q2 + p2 q1 = 0 ⇒ q2 ( p1 − q1 ) = 0,

implying either q2 other.

= 0 = p2 or p1 = q 1. Hence either we need both pand qreal, or pand qare complex conjugates of each

EXAMPLE

Find a quadratic whose roots are Let the quadratic be x2

2 and 5.

+ bx + c, where we wish to find b and c. Then Vieta's formula tells us that b = −(2 + 5) = −7,

Therefore the desired quadratic is x2

c = 2 × 5 = 10.

− 7x + 10. □

EXAMPLE

Find a quadratic whose roots are

3 + 2i and 3 − 2i.

Let the desired quadratic be x 2 +

bx + c, where we wish to find band c. Then Vieta's formula tells us

b = − [(3 + 2i) + (3 − 2i)] = −6, So the desired quadratic is x2

c = (3 + 2i ) (3 − 2i ) = 13.

− 6x + 13. □

Note: Since the roots were complex conjugates of each other, the coecients of the quadratic became real.

EXAMPLE

Solve the system of equations

a + b = 7,

ab = 10.

By Vieta's Formula, we know that a and b are the roots of the equation

x 2 − 7x + 10 = 0. Since we can factorize it as (x −

2)(x − 5) = 0, we get that {a, b} = {2, 5}. □

Generalization to Higher Degree Polynomials Consider a quadratic equation with complex coecients and roots

r1, r2 :

a2 x2 + a1 x + a 0 = a2 (x − r 1 )(x − r2 ). https://brilliant.org/wiki/vietas-formula/

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By comparing coecients, we can see that

r1 + r2 = − aa12 ,

r1 r2 =

a0 a2

.

This gives a relationship between the roots of a polynomial and the coecients of the polynomial. Generalizing this idea for a polynomial of degree n, we have the following formula: D EFI NITI ON

Vieta's Formula:

= a n xn + an−1 xn−1 + ⋯ + a0 be a polynomial with complex coecients and degree n, having complex roots r n, r n−1, … , r 1. Then for any integer 0 ≤ k ≤ n,

Let P (x)

∑

ri 1 ri2 ⋯ ri k = (−1)k

1≤i 1...