There are clear relationships with the roots of a polynomial and the coefficients of the latter.

x^{2} - 3x - 10 = 0,

**The roots are, x = 5 or x = -2.**

Let's look at the coefficients now:

b/a = -3/1 = -3; c/a = -10/1 = -10

From the roots,

5 + (-2) = 3; 5 x -2 =-10

So, if the roots of ax^{2} + bx + c = 0, are α and β,

α + β = -b/a and αβ = c/a

In general, if ax^{2} + bx + c= a(x - α)(x - β),

Sum of Roots, α + β = -b/a and Product of Roots, αβ = c/a

*Proof*

ax^{2} + bx + c= a(x - α)(x - β) = a(x^{2} - (α + β)x + αβ) = ax^{2} - a(α + β)x + aαβ

Making the coefficients of x and the constants on both sides equal,

b = -a(α + β) => (α + β) = -b/a

c = aαβ => αβ = c/a

**E.g.1**

If x^{2} - 4x - 12 = 0, find the values of α + β, αβ (α+β)/αβ and α^{2} + β^{2}, without solving the equation.

1/α + 1/β = (α+β)/αβ = 4/-12 = -1/3

(α+β)

α

= 16 - 2 x (-12) = 16 + 24 = 40.

**E.g.2**

If 2x^{2} - 5x - 12 = 0, find the values of α + β, αβ and (α+β)/αβ, without solving the equation.

1/α + 1/β = (α+β)/αβ = -2.5/6 = -5/12

**E.g.3**

If px^{2} + qx + r = 0, find the values of p, q and r, if α = -2/3 and β= 5/6.

So, (x + 2/3)(x - 5/6) = 0

x

18x

p = 18; q = -3; r = -10

= a[(x

= ax

α + β + γ = -b/a; αβ + αγ + βγ = c/a; αβγ = -d/a

In general, if ax

Sum of Roots, α + β + γ = -b/a; Sum of products of pairs of roots,αβ + αγ + βγ = c/a; Product of Roots, αβγ = -d/a

**E.g.1**

If α β and γ are the roots of the polynomial, 2x^{3} - 5x^{2} -9x + 18, find α + β + γ, αβ + αγ + βγ, αβγ and 1/α + 1/β + 1/γ, without finding the roots.

α + β + γ = -b/a = -(-5/2) = 2.5

αβ + αγ + βγ = c/a = -9/2 = -4.5

αβγ = -d/a = -(18/2) = -9

1/α + 1/β + 1/γ = (αβ + αγ + βγ)/(αβγ) = -4.5/-9 = 1/2

If the roots of the cubic polynomial are complex, they are conjugates. The following question is based on a combination of complex and real roots.

**E.g.2**

If the roots of ax^{3} + bx^{2} +cx + d are α = 2 + i, β = 2 - i and γ = 3, find the values of a, b, c and d.

ax^{3} + bx^{2} +cx + d = (x - 2 - i)(x -2 + i)(x - 3)

= (x^{2} - 2x + ix -2x + 4 - 2i -ix + 2i + 1)(x - 3)

= x^{3} - 7x^{2} + 17x - 15

= x^{3} - 7x^{2} + 17x - 15

Making the coefficients on both sides equal,

a = 1; b = -7; c = 17; d = -15

= a[(x

= ax

α + β + γ + δ = -b/a; αβ + αγ + αδ + βγ + βδ + γδ = c/a; αβγ + αβδ + βδγ + αδγ = -d/a; αβγδ = e/a

In general, if ax

Sum of Roots, α + β + γ + δ = -b/a; Sum of products of pairs of roots,αβ + αγ + αδ + βγ + βδ + γδ = c/a; Sum of Product of Triplets of Roots = αβγ + αβδ + βδγ +αδγ = -d/a ; Product of Roots, αβγδ = e/a

**E.g.1**

If x^{4} - x^{3} - 7x^{2} + x + 6 = 0, find α + β + γ + δ, αβγδ and 1/α + 1/β + 1/γ + 1/δ.

α + β + γ + δ = -b/a = 1

αβγδ = e/a = 6

1/α + 1/β + 1/γ + 1/δ = (αγ + αδ + βγ + βδ + γδ)/(αβγδ) = (-c/a)/(e/a)

= -7/6

**E.g.2**

If x^{4} - 5x^{3} + cx^{2} + dx + e = 0 and the two complex roots of the function are 2 ±i3. Find the other roots and the values of c and d.

Let α = 2 + i3, β = 2 - i3 and the other roots be γ and δ.

Since α + β + γ + δ = -b/a => 2 + i3 + 2 -i3 + γ + δ = 5

γ + δ = 1

αβγδ = e/a => αβγδ = -78

(2 + i3)(2 - i3)γδ = -78

13γδ = -78 => γδ = -6

Solving the two equations, we get γ = 3; δ =-2

αβ + αγ + αδ + βγ + βδ + γδ = c/a

13 + (6 + i9) + (-4 -i6) + (6 - 9i) + (-4 + 6i) - 6 = c/1 => c = 11

αβγ + αβδ + βδγ + αδγ = -d/a

-26 + 39 + (-12 - i18) + (-12 + i18) = -d/a

-11 = d/1 => d = 11

Suppose the roots of a quadratic polynomial are α and β. If their values are changed to, say, α + 1 or β + 1, then, there is going to be a corresponding change in the original polynomial - and a corresponding transformation of the graph too. In this section, the transformation of the graph and the corresponding formation of the new polynomial are explored in detail.

**E.g.1**

x^{2} - x - 6 = 0

α + β = -b/a = 1

αβ = c/a = -6

α = 3; β = -2

Now let's transform the two roots - a **translation** by 1 in the **negative x-axis**: α + 1 = 4; β + 1 = -1;
So, the new roots are, x = 4 and x = -1

Now let's form the new polynomial:

New α = 4; new β = -1;

New α + β = 3; new αβ = -4

Since α + β = -b/a and αβ = c/a, if a =1, the new polynomial is derived as follows:

In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:

w^{2} - 3w - 4 = 0

**E.g.2**

x^{2} - x - 6 = 0

α + β = -b/a = 1

αβ = c/a = -6

α = 3; β = -2

Now let's transform the two roots - **stretching** by a factor 1/2 **parallel to x-axis**: α/2 = 3/2; 2β = -1;
So, the new roots are, x = 3/2 and x = -1

Now let's form the new polynomial:

New α = 3/2; new β = -1;

New α + β = 1/2; new αβ = -3/2

Since α + β = -b/a and αβ = c/a, the new polynomial is derived as follows:

In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:

w^{2} - 1/2 w - 3/2 = 0

4w^{2} -2w - 6 = 0

**E.g.3**

x^{2} - x - 6 = 0

α + β = -b/a = 1

αβ = c/a = -6

α = 3; β = -2

Now let's transform the two roots - **reflecting** in the **y-axis**: -1 x α = -3; -1 x β = 2;
So, the new roots are, x = -3 and x = 2

Now let's form the new polynomial:

New α = -3; new β = 2;

New α + β = -1; new αβ = -6

Since α + β = -b/a and αβ = c/a, the new polynomial is derived as follows:

In order to minimize the confusion, it is recommended to use a new variable for the new polynomial, instead of x. Using w as the new variable, the new polynomial takes the following form:

w^{2} + w - 6 = 0

You may practise the above interactively with the following applet; just move the slider and see the corresponding change in the polynomial and the graph. Please select one transformation at a time to avoid the text and graphics being blurred.

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7^{th} edition in print.