A quadratic equation (shorten as quadratic) is one in which the highest power of the unknown quantity is 2. We’ll outline different ways of solving quadratic equations, including the most general method of using the quadratic equation formula. Working through the extensive range of various examples presented here, you will be prepared to solve quadratic equations of any difficulty.

Explore the various forms of quadratic equations here!

Defining the Quadratic Formula

We define a quadratic equation to be an equation that can be put in the form:

a x^2 + b x + c = 0

Where x represents an unknown variable, and parameters a, b, and c are the coefficients of the equation. This form is referred to as the standard form of a quadratic equation. The quadratic coefficient a is not equal to 0, because if a = 0, then the equation is linear, not quadratic.

To solve the quadratic equation means to find the value of the unknown. The value of the unknown is called the root of the equation.

There are three methods of solving quadratic equations. These are:

1) by factoring (if possible)

2) by completing the squire

3) by using the quadratic equation formula

We investigate all these methods in more detail.

Solving Quadratic Formulas by Factoring

If the quadratic expression in a quadratic equation can be factorized, the equation can be solved using the zero product property, which states that for any algebraic expressions A and B,

A B = 0 quad text{if and only if} quad A = 0 quad text{or} quad B = 0

In other words, a product can be zero if and only if one of its factors is zero.

If the quadratic expression can be factorized this provides the most simple way of solving the quadratic. For example, consider equation:

6 x^2 = 3 x

We can easily observe that it can be put in the standard form for a quadratic equation and factorized:

6 x^2 - 3 x = 3 x ( 2 x - 1 ) = 0

Due to the zero product property, we can determine two roots:

x_1 = 0 ,, quad quad x_2 = dfrac{1}{2}

Let’s consider a more complicated example:

6 x^2 - 5 x + 1 = 0

We can resolve this quadratic if we notice that in can be written in the form:

6 x^2 - 5 x + 1 = 6 x^2 - 3 x - 2 x + 1 = 3 x ( 2 x - 1 ) - ( 2 x - 1 ) = 0

Now, factoring is easy:

( 3 x - 1 )( 2 x - 1 ) = 0

Which results in two roots:

x_1 = dfrac{1}{3} ,, quad quad x_2 = dfrac{1}{2}

Sometimes the process of factoring a quadratic equation requires a considerable effort of trial and error and guess-and-check. Large amounts of quadratic equations that appear in practice cannot be solved by factoring.

Solving Quadratic Formulas by Completing the Square

An expression such as:

x^2 or (x-p)^2 ,

Where x is unknown variable and p is some real number, is called a perfect square. If a quadratic has a form of a perfect square, it can be resolved by using the fact that

text{if} quad x^2 = k,, quad text{then} quad x_1 = sqrt{k} quad text{and} quad x_2 = - sqrt{k}

Analogously:

text{if} quad ( x - p )^2 = k,, quad text{then} quad x_1 = p + sqrt{k} quad text{and} quad x_2 = p - sqrt{k}

Consequently if a quadratic equation can be permuted so that one side of the equation is a perfect square and the other side is a number, then the solution of the quadratic is readily obtained by taking the square roots of each side as in the above expressions. The process of permuting one side of a quadratic equation into a perfect square before solving is called “completing the square’’. The key to this method is the identity

( x + t )^2 = x^2 + 2 t x + t^2

Thus, in order to make the quadratic expression  x^2 + bx  into a perfect square, we have to perform the following transformation:

x^2 + b x = x^2 + b x + dfrac{b^2}{4} - dfrac{b^2}{4} = left( x + dfrac{b}{2} right)^2 - left( dfrac{b}{2} right)^2

Let’s see a few examples of how it works. Consider the quadratic:

x^2 + 4 x = - 1

To complete the square, we add 4 to both sides of this equation:

x^2 + 4 x + 4 = 4 - 1 = 3

Using the fact:

x^2 + 4 x + 4 = ( x + 2 )^2

We can write this equation in the form of:

( x + 2 )^2 = 3

Which can be easily resolved:

x_1 = - ,2 + sqrt{3} ,, quad quad x_2 = - ,2 - sqrt{3}

The quadratic coefficient is not necessarily equal to unity in this method. For instance we can solve:

2 x^2 + 5 x - 3 = 0

First, make the coefficient of the  x^2  term unity. This is achieved by dividing throughout by 2:

x^2 + dfrac{5}{2} x = dfrac{3}{2}

The coefficient of x is 5/2. Thus, to complete the square we add  (5/4)^2  to both sides of this equation:

x^2 + dfrac{5}{2} x + left( dfrac{5}{4} right)^2= dfrac{3}{2} + left( dfrac{5}{4} right)^2= dfrac{49}{16}

Now we see the left hand side of this equation is a perfect square:

left( x + dfrac{5}{4} right)^2= dfrac{49}{16}

Both roots of this equation can be easily determined:

x_1 = - ,dfrac{5}{4} + sqrt{dfrac{49}{16}} = - ,dfrac{5}{4} + dfrac{7}{4} = dfrac{1}{2}

And:

x_2 = - ,dfrac{5}{4} - sqrt{dfrac{49}{16}} = - ,dfrac{5}{4} - dfrac{7}{4} = - ,3

Hence:

x_1 = 1/2 and x_2 = - 3

These are the two roots of the quadratic:

2 x^2+ 5 x - 3 = 0

When the methods mentioned above do not apply (or even if they do), a quadratic equation can be solved using the quadratic equation formula.

Solving Quadratic Formulas by Using the Quadratic Equation Formula

Let’s see what the method of completing the square gives in the most general case of the quadratic equation given by:

a x^2 + b x + c = 0

Dividing this expression by a gives:

x^2 + dfrac{b}{a} ,x+ dfrac{c}{a} = 0

Which can be written in the form:

x^2 + dfrac{b}{a} ,x= -, dfrac{c}{a}

Now we add to each side of the equation term  (b/2a)^2  to make the left hand side a perfect square:

x^2 + dfrac{b}{a} ,x + left( dfrac{b}{2a} right)^2 = left( dfrac{b}{2a} right)^2 -, dfrac{c}{a}

Thus we have:

left( x + dfrac{b}{2a} right)^2 = dfrac{b^2}{4a^2} -, dfrac{c}{a} = dfrac{b^2 - 4ac}{4 a^2}

Taking the square roots of both sides gives:

x + dfrac{b}{2a} = sqrt{ dfrac{b^2 - 4ac}{4 a^2} } = dfrac{ pm sqrt{b^2 - 4ac} }{2a}

Hence general solution of quadratic can be presented in the form:

x_{1,2} = dfrac{ - b pm sqrt{b^2 - 4ac}}{2a}

Which is known as the quadratic equation formula (sometimes shorten as quadratic formula).

The expression:

D = b^2 - 4 a c

is called the discriminant of the quadratic equation

a x^2 + b x + c = 0 .

As we see from the quadratic equation formula, solutions of this equation can be categorized as follows:

– If  D = b^2 - 4 a c < 0, then the quadratic equation has no (real) solutions.

– If  D = b^2 - 4 a c = 0, then the quadratic equation has one solution:  x = -, dfrac{b}{2a}

– If  D = b^2 - 4 a c > 0, then the quadratic equation has two solutions, given by the quadratic formula:  x_{1,2} = dfrac{ - b pm sqrt{D}}{2a}

The behavior of quadratic expressions with different discriminants are illustrated below:

We note the graph of a quadratic function:

f(x) = a x^2 + b x + c

Always produces a shape called a parabola.

So quadratic equation of the form:

a x^2 + b x + c = 0

This may be solved graphically by plotting the graph:

y = a x^2 + b x + c

And noting the points of intersection on the x-axis (where y = 0). The number of roots of the quadratic depends on how many times the curve intercepts the x-axis. As we can see from the image, the quadratic equation with D< 0 has no roots because corresponding parabola does not overlap the x-axes. If D< 0, parabola touches x-axes at one point. Finally, in the case D> 0, both branches of the parabola intercepts x-axes, which means quadratic equation has two roots.

Consider examples of applying the quadratic equation formula. We begin from the equation:

x^2 + 6 x + 13 = 0

Comparing this quadratic with quadratic of standard form:

a x^2 + b x + c = 0

We see that in this case:

a = 1 ,, quad b = 6 quad text{and} quad c = 13

Calculate the discriminant:

D = b^2 - 4 a c = 6^2 - 4cdot 1 cdot 13 = 36 - 144 = - ,16

As the discriminant is negative we conclude that the quadratic equation under investigation has no real roots.

Quadratic function:

f(x) = 4 x^2 - 12 x + 9

This has a zero discriminant, which can be easily verified:

a = 4 ,, quad b = - 12 quad text{and} quad c = 9

And:

D = b^2 - 4 a c = ( -12)^2 - 4cdot 4 cdot 9 = 144 - 144 = 0

Thus the quadratic equation:

4 x^2 - 12 x + 9 = 0

This has only one root which can be determined by the formula:

x = -, dfrac{b}{2a} = -, dfrac{(-12)}{2cdot 4} = dfrac{3}{2}

Now let’s solve:

x^2 + 2 x - 8 = 0

Comparing this quadratic with the quadratic of the standard form we see that:

a = 1 ,, quad b = 2 quad text{and} quad c = -, 8

The calculation of the discriminant gives us:

D = b^2 - 4 a c = 2^2 - 4cdot 1 cdot (-8) = 4 + 32 = 36

Since the discriminant is positive, we expect to obtain two different solutions here. Applying the quadratic equation formula we obtain:

x_{1} = dfrac{ - b + sqrt{D}}{2a} = dfrac{ - 2 + sqrt{36}}{2} = 2

And:

x_{2} = dfrac{ - b - sqrt{D}}{2a} = dfrac{ - 2 - sqrt{36}}{2} = - ,4

Thus:

x_1 = 2  and  x_2 = - 4

These are the two roots of the equation.

x^2 + 2 x - 8 = 0 .

Finally, consider the equation:

dfrac{x + 1}{x - 1} = x - 3

Multiplying both sides by (x - 1) we can easily transform this equation into a quadratic of standard form:

(x - 3)(x - 1) = x + 1

This results in:

x^2 - 4 x + 3 = x + 1

and

x^2 - 5 x + 2 = 0

Coefficients of this quadratic equation:

a = 1 ,, quad b = - 5 quad text{and} quad c = 2

This can be now used to determine the discriminant:

D = b^2 - 4 a c = 5^2 - 4cdot 1 cdot 2 = 25 - 8 = 17

This appears to be positive. According to the quadratic equation formula this means the quadratic equation has two roots, which can be expressed as:

x_{1} = dfrac{ - b + sqrt{D}}{2a} = dfrac{ 5 + sqrt{17}}{2}  and  x_{2} = dfrac{ - b - sqrt{D}}{2a} = dfrac{ 5 - sqrt{17}}{2}

Wrapping Everything Up

In conclusion, applying quadratic equation formula is the most general way of solving quadratics, which can be applied when more simple methods of factoring and completing the square do not work. We hope you have mastered the quadratic formula and are ready to use it for solving quadratic equations of any difficulty.

Let’s put everything into practice. Try this Algebra practice question:

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