Solving quadratic equations

There are three main ways to solve quadratic equations:

factorising
using the quadratic formula
graphically

Let's look at all three ways, one at a time.

Example one

Find the solutions to the equation \({x^2}\) + 9\({x}\) + 14 = 0

Method one: factorising

We can factorise the equation above to give (\({x}\) + 2)(\({x}\) + 7) = 0

For tips on how to do this, look at Factorising quadratics and Factorisation of further quadratics in this guide.

For (\({x}\) + 2)(\({x}\) + 7) to equal 0 either the first or second bracket must be equal to 0. This means that either \({x}\) is -2 or \({x}\) is -7.

If \({x}\) = -2 we have: (0) × (5) = 0.

If \({x}\) = -7 we have: (-5) × (0) = 0.

So the solutions of the equation are \({x}\) = -2 and \({x}\) = -7.

Method two: the quadratic formula

We can solve expressions of the form a\({x^2}\) + b\({x}\) + c = 0 using the quadratic formula:

\({x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}\)

To use this formula we simply substitute the values from our equation and calculate the result.

We have \({x^2}\) + 9\({x}\) + 14 = 0 so:

\({a}\) = 1 (\({a}\) is 1 because we have one lot of \({x^2}\))

\({b}\) = 9

\({c}\) = 14

When substituted into the quadratic formula we have:

\({x = \frac{ - 9 \pm \sqrt {9^2 - 4~\times1\times14} }{{2}~\times~1}}\)

\({x = \frac{ - 9 \pm \sqrt {81 - 56} }{{2}}}\)

\({x = \frac{ - 9 \pm \sqrt {25} }{{2}}}\)

\({x = \frac{ - 9 \pm 5 }{{2}}}\)

Now we split the equation into two because of the ± sign.

\({x = \frac{ - 9 + 5 }{{2}}}~and~{x = \frac{ - 9 - 5 }{{2}}}\)

\({x = \frac{ - 4 }{{2}}}~and~{x = \frac{ - 14 }{{2}}}\)

Which finally leaves:

\({x = -2}~and~{x = -7}\) as before.

Method three: graphically

Plot the graph of \({y}\) = \({x^2}\) + 9\({x}\) + 14

Where the graph crosses the line \({y}\) = 0 we will have our solutions.

As the graph is quadratic we can see that the line crosses the \({x}\)-axis twice, once at \({x}\) = -7 and once at \({x}\) = -2. Therefore the solutions are \({x}\) = -7 and \({x}\) = -2.

Example two

Solve the equation: \({x^2}\) – 25² = 0

Method one: factorising

\({x^2}\) – 25² = (\({x}\) - 25)(\({x}\) + 25) = 0

Therefore either (\({x}\) – 25) or (\({x}\) + 25) = 0

This means that either \({x}\) = 25 or \({x}\) = –25

So our solutions are \({x}\) = 25 and \({x}\) = –25

Method two: the quadratic formula

We have \({x^2}\) – 25² = 0 giving us the following values to substitute into the quadratic formula:

\({a}\) = 1

\({b}\) = 0

\({c}\) = –25² = –625

\({x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}\)

So substituting in we have:

\({x = \frac{ \pm \sqrt {-4 \times -625} }{{2}}}\)

\({x = \frac{ \pm \sqrt {2500} }{{2}}}\)

\({x = \frac{ \pm {50} }{{2}}}\)

\({x = -25}~and~{x = 25}\)

Method three: graphically

Plotting the graph of \({y}~=~{x^2}\) – 25²

The graph of y = x squared - 25 squared.

From the graph we can see that there are two points of intersection, one at –25 and one at 25.

Our solutions are \({x}\) = 25 and \({x}\) = –25.

Which method should I choose?

It is normally quite difficult to accurately plot and read off the solutions to quadratic equations by using a graphical method – so only do this if the question specifically asks for it.

Some quadratics have solutions that are not whole numbers, these would be incredibly difficult to factorise so the quadratic formula is used.

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Quadratic expressions - Intermediate & Higher tier – WJECSolving quadratic equations