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# Sum and product of the roots of a quadratic equation

We learned on the previous page (The Quadratic Formula), in general there are two roots for any quadratic equation ax^2+ bx + c = 0. Let's denote those roots alpha and beta, as follows:

alpha=(-b+sqrt(b^2-4ac))/(2a) and

beta=(-b-sqrt(b^2-4ac))/(2a)

## Sum of the roots α and β

We can add alpha and beta as follows:

alpha + beta =(-b+sqrt(b^2-4ac))/(2a)+(-b-sqrt(b^2-4ac))/(2a)

 =(-2b+0)/(2a)

 =-b/a

## Product of the roots α and β

We can multiply alpha and beta as follows. First, recall that in general,

(X+Y)(X-Y) = X^2 - Y^2 and

(sqrt(X))^2 = X

We make use of these to obtain:

alpha xx beta = (-b+sqrt(b^2-4ac))/(2a) xx (-b-sqrt(b^2-4ac))/(2a)

 =((-b)^2 - (sqrt(b^2-4ac))^2)/(2a)^2

 =(b^2 - (b^2 - 4ac) )/(4a^2)

 =(4ac)/(4a^2)

 =c/a

## Summary

The sum of the roots alpha and beta of a quadratic equation are:

alpha + beta = -b/a

The product of the roots alpha and beta is given by:

alpha beta = c/a

It's also important to realize that if alpha and beta are roots, then:

(x-alpha)(x-beta)=0

We can expand the left side of the above equation to give us the following form for the quadratic formula:

x^2 - (alpha+beta)x + alpha beta = 0

Let's use these results to solve a few problems.

### Example 1

The quadratic equation 2x^2- 7x - 5 = 0 has roots alpha and beta. Find:

(a) alpha + beta

(b) alpha beta

(c) alpha^2 + beta^2

(d) 1/alpha + 1/beta

For the expression 2x^2- 7x - 5, we have:

a=2

b=-7

c=-5

(a) We learned just now that alpha + beta = -b/a so in this example,

alpha + beta = -((-7))/2 = 3.5

(b) We know alpha beta = c/a so in this example,

alpha beta = (-5)/2 = -2.5

(c) For alpha^2 + beta^2, we need to recall that

(alpha + beta)^2 = alpha^2 + 2alpha beta + beta^2.

Solving this for alpha^2 + beta^2 gives us:

alpha^2 + beta^2 = (alpha + beta)^2 - 2alpha beta.

We've already found the sum and product of alpha and beta, so we can substitute as follows:

alpha^2 + beta^2 = (3.5)^2 - 2xx(-2.5) = 17.25.

(d) We add our fractions 1/alpha + 1/beta as follows:

1/alpha + 1/beta = (beta + alpha)/(alpha beta) = (alpha + beta)/(alpha beta)

We know the sum (top) and product (bottom), so we can simply write:

1/alpha + 1/beta = (alpha + beta)/(alpha beta) = 3.5/(-2.5) = -1.4

### Example 2

Find the quadratic equation with roots α and β given αβ = 2 and α2β2 = 3.

We'll set up a system of two equations in two unknowns to find alpha and beta.

Remembering the difference of squares formula, we have

α2β2 = (α + β)(α β)

From the question we know α2β2 = 3, so this gives us:

3 = (α + β)(α β)

The question says αβ = 2, which we can substitute into the right hand side, giving:

3 = 2(α + β)

This gives:

(alpha + beta) = 3/2

Using αβ = 2 again, we add it to the above line, giving:

2 alpha = 3/2 + 2 = 7/2

So alpha = 7/4

Since (alpha + beta) = 3/2 then beta = 3/2 - alpha, giving us beta = -1/4.

We substitute these values into the expression x^2 - (alpha+beta)x + alpha beta = 0 giving:

x^2 - (3/2)x + (7/4)(-1/4) = 0

x^2 -3/2 x -7/16=0

So the required quadratic equation is:

x^2 -3/2 x -7/16 = 0

We multiply throughout by 16 to tidy it up:

16x^2 - 24x - 7 = 0`

Let's now go on to learn how the graph of a quadratic function is a parabola: 4. The Graph of the Quadratic Function

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