Skip to main content

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


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`


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:


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) 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`

Get the Daily Math Tweet!
IntMath on Twitter

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

Search IntMath

Online Algebra Solver

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

Algebra Lessons on DVD

Math videos by

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!

See the Interactive Mathematics spam guarantee.