# 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`

Answer

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 *α* − *β* = 2 and *α*^{2} − *β*^{2} = 3.

Answer

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

α^{2}−β^{2}= (α+β)(α−β)

From the question we know *α*^{2} − *β*^{2} = 3, so this gives us:

3 = (

α+β)(α−β)

The question says *α* − *β* = 2, which we can substitute into the right hand side, giving:

3 = 2(

α+β)

This gives:

`(alpha + beta) = 3/2`

Using *α* − *β* = 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 + (3/2)(-1/4) = 0`

`x^2 -3/2 x -3/8`

So the required quadratic equation is:

`x^2 -3/2 x -3/8 = 0`

We multiply throughout by `8` to tidy it up:

`8x^2 - 12x - 3 = 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

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!