Quadratic equations

A quadratic equation is an equation with one unknown that is raised to the power 2. If the equation contains the unknown raised to a higher power than 2, it is no longer a quadratic equation.

Describing a quadratic equation

A quadratic equation can always be written in this form:

$a x^{2} + b x + c = 0$

The values a, b and c are real numbers and a is not zero. The parts of the equation have their own names:

ax² is called the quadratic term,
bx is called the linear term,
c is called the constant term.

Every quadratic equation can be rearranged into the form ax² + bx + c = 0 by doing the same to both sides or by other manipulations. An example of a quadratic equation is

$3 x^{2} - 6 x + 8 = 0.$

A harder question is whether this is also a quadratic equation:

$(x + 1) \cdot (x + 2) = 7.$

There is no x² in sight and the right-hand side is not zero. The right-hand side is easy to deal with: we subtract 7 from both sides, which gives

$(x + 1) \cdot (x + 2) - 7 = 0.$

Still no x². Let’s expand the brackets:

$x^{2} + 2 x + x + 2 - 7 = 0$

Now there is an x². After collecting like terms we finally get the form ax² + bx + c = 0:

$x^{2} + 3 x - 5 = 0.$

What about the next example? Is it a quadratic equation, and can it be rearranged into that form?

$x^{2} + 3 x + \frac{4}{x} = 0$

The answer is that it cannot. If we multiply the whole equation by x to get rid of the variable in the denominator, we get

$x^{3} + 3 x^{2} + 4 = 0.$

Now there is an x³, which cannot appear in a quadratic equation.

There are special types of quadratic equations, depending on the values of the coefficients a, b and c.

Finding the coefficients

For the later sections it is absolutely vital that you can pick out the coefficients a, b and c. Let’s start with a simple example. Find the coefficients of 3x² + 4x + 7 = 0. This one is easy: a = 3, b = 4 and c = 7.

What about 3x² − x + 2 = 0? The values of a and c are clear, a = 3 and c = 2, but what about b? It would be a mistake to say b = 1. Imagine there is a 1 in front of the x, like this: 3x² − 1x + 2 = 0. Now it is easier to see that b = −1.

Another interesting example: x² − 3 = 0. Again the coefficient b can cause trouble; the others are clear: a = 1 and c = −3. What is b? There is no linear term at all in this equation, which is the same as writing x² + 0x − 3 = 0. So b = 0.

A trickier example:

$5 x^{2} + 3 x + π \cdot x = 0$

This equation has two traps. There is no constant term, so c = 0. The quadratic term is clear, so a = 5. But what about that wretched b? We can rearrange the equation a little by taking out the common factor x from the two terms:

$5 x^{2} + (3 + π) x = 0$

The coefficient b is again whatever is “in front of” the variable x, so the whole bracket is the coefficient: b = (3 + π). A similar equation:

$- x^{2} + 4 x + 1 - \sqrt{2} = 0$

The first two coefficients are clear: a = −1 and b = 4. What is the constant term? It is all the remaining terms that do not contain x, so $c = 1 - \sqrt{2}$ . Don’t be afraid of the square root; it doesn’t bite, it is just an ordinary number.

And what are the values for this equation? (x + 1)(x + 2) = −2x. First we must rearrange it into the form ax² + bx + c = 0. Expanding the brackets gives

$x^{2} + 3 x + 2 = - 2 x$

and finally we add 2x to both sides to get zero on the right-hand side:

$x^{2} + 5 x + 2 = 0.$

Now the values are clear: a = 1, b = 5 and c = 2.

Quadratic equations with no x term

If b = 0, the equation has no x term and is solved much like a linear equation. Such an equation has the form

$a x^{2} + c = 0$

Because b = 0, the whole linear term bx drops out of the equation, since 0x = 0. We solve the equation by first subtracting the constant term, that is, subtracting c from both sides. This gives:

$\begin{array}{rcl} a x^{2} + c & = & 0 / - c \\ a x^{2} & = & - c \end{array}$

Next we divide both sides by a. We get

$x^{2} = \frac{- c}{a}$

And finally we take the square root of both sides. For this to work, the fraction on the right-hand side must not be negative, because you cannot take the square root of a negative number. The result is:

$x_{1} = \sqrt{\frac{- c}{a}}, x_{2} = - \sqrt{\frac{- c}{a}}$

Notice that we get two solutions, one positive and one negative. This is easier to see in an example. Let’s solve the equation

$2 x^{2} - 8 = 0.$

Following the method, we first subtract the constant term c. Watch out: here c is negative, so we are subtracting a negative number, which means adding a positive one. We add +8 to both sides (the aim is to get the 8 onto the right-hand side). This gives:

$2 x^{2} = 8$

Next we divide both sides by a, which is 2. This gives:

$x^{2} = 4$

Now we take the square root.

$x_{1} = 2, x_{2} = - 2$

Why is minus 2 a solution as well? Because minus 2 times minus 2 is 4. We can also look at the solution graphically: the graph of the function 2x² − 8 crosses the x-axis exactly at the points x₁ = −2 and x₂ = 2.

Graphical solution of the equation 2x^2-8=0

Solve the equation

$80 x^{2} - 5 = 0.$

We add 5 to both sides:

$80 x^{2} = 5$

Divide by 80:

$x^{2} = \frac{5}{80}$

Cancel the fraction:

$x^{2} = \frac{1}{16}$

Take the square root:

$x_{1} = \frac{1}{4}, x_{2} = - \frac{1}{4}$

Quadratic equations with no constant term

Another special case arises when the constant term c is zero. The equation then has the form

$a x^{2} + b x = 0$

This type of equation is solved by taking out the common factor x:

$x (a x + b) = 0.$

The result is now fairly clear. Again there are two roots, but one of them is always zero. The whole left-hand side is a product of two expressions: x and the bracket (ax + b). When is this product zero? When at least one of the two expressions is zero. So the left-hand side is zero in two cases:

$x = 0 or a x + b = 0.$

We already have the first solution: it is zero. We get the other by solving the linear equation ax + b = 0. From the article on linear equations we know that this equation has exactly one solution:

$x = - \frac{b}{a}$

So we can now write down the solution of the whole quadratic equation ax² + bx = 0:

$x_{1} = 0, x_{2} = - \frac{b}{a} .$

Example:

$6 x^{2} + 3 x = 0$

Take out x:

$x (6 x + 3) = 0$

The first solution is x = 0. We find the second by solving the linear equation 6x + 3 = 0:

$x = - \frac{b}{a} = - \frac{3}{6} = - \frac{1}{2} .$

Write down the result:

$x_{1} = 0, x_{2} = - \frac{1}{2} .$

Again we can look at the graph of the function, which crosses the x-axis exactly at these points:

Graph of the function 6x^2+3x

Other ways of solving

These were the basic techniques for special types of quadratic equations. A general quadratic equation is solved using the discriminant. Other types of equations include the quadratic equation with a parameter.

Calculator

Enter the coefficients a, b and c of the general equation ax² + bx + c = 0, and the calculator will work out the discriminant and the roots. You can also find it on its own as the quadratic equation calculator.

Chapters: Quadratic equations, The discriminant, Quadratic equations with a parameter, Quadratic equations with complex solutions

« Previous: Linear equations with a parameter

Next: The discriminant »