Solving Quadratics by Factorising

Example 1:

Solve the following quadratics by factorising:

a)

Step 1: Factorise the quadratic expression on the left side of the equation.

First, we look at what multiplies to give Since we write in both brackets.

Next, we look for two numbers that multiply to give and add to give We list the factor pairs of

The pairs are:

and

and

and

and

and

Out of these pairs, only and can make To get we do Therefore, the quadratic factorises to

Now, we have a product of two factors that equal so one or both factors must be We have:

or

Step 2: Solve each linear equation for to find the roots.

For the first equation:

Subtracting from both sides to isolate

For the second equation:

Adding to both sides to isolate

Therefore, the roots of the quadratic equation are or

b)

Step 1: Factorise the quadratic expression on the left side of the equation.

The quadratic expression is an example of the difference of two squares principle. Recall that

Using this principle, we factorise the quadratic.

So, setting each factor equal to we have

or

Step 2: Solve each linear equation for to find the roots.

For the first equation:

Subtracting from both sides,

Dividing both sides by

For the second equation:

Adding to both sides,

Dividing both sides by

Therefore, the roots of the quadratic equation are or

Note: Alternatively, we could solve for here by simply rearranging the equation for

First, we add to both sides.

Then, we divide both sides by

Finally, we square root both sides to isolate Remember that when we square root, it will produce a positive and a negative solution.

Therefore, the solutions are or

c)

Step 1: Factorise the quadratic expression on the left side of the equation.

First, we look at what multiplies to give The only option is and so we write these in the brackets.

We now look for two numbers that multiply to give We list the factor pairs of

The pairs are:

and

and

Usually, we should first choose the pair whose values are closest together, which will be and We write in both brackets.

We now apply the OI-OI method to check whether we can obtain the correct middle term, which is using this arrangement.

Here, the outer product is and the inner product is We can make by doing Therefore, the quadratic factorises to

So, setting each factor equal to we have,

or

Step 2: Solve each linear equation for to find the roots.

For the first equation:

Adding to both sides,

Dividing both sides by

For the second equation:

Subtracting from both sides,

Therefore, the roots of the quadratic equation are and

Example 2:

The area of this triangle is

a) Show that

Step 1: Use the formula for the area of a triangle and our lengths to form an equation.

Recall that the area of a triangle is given by

Base Height

Here, we have a base of cm and a height of cm. So,

We can multiply both sides by to cancel out the on the left side.

Step 2: Expand the brackets using FOIL and rearrange the equation to get it in the form required by the question.

Subtracting from both sides so that one equals

b) Work out the value of

We need to solve the following quadratic equation found in part (a).

Step 1: Factorise the quadratic expression on the left side of the equation.

First, we look at what multiplies to give The only option is and so we write these in the brackets.

We now look for two numbers that multiply to give We list the factor pairs of

The pairs are:

and

and

and

First, we should choose the pair whose values are closest together, which will be and We write in both brackets.

We now apply the OI-OI method to check whether this arrangement gives the middle term, which is

Here, the outer product is and the inner product is However, we cannot make using these products, so and are not a suitable pair.

We therefore try the next closest factor pair, which is and We now decide which bracket the and should go in. The options are

or

Once again, we apply the OI-OI method to check whether any of these arrangements give the middle term, which is

Here, the outer product is and the inner product is We can make by doing Therefore, the quadratic factorises to

So, setting each factor equal to we have,

or

Step 2: Solve each linear equation for to find the roots.

For the first equation:

Adding to both sides,

Dividing both sides by

For the second equation:

Subtracting from both sides,

Therefore, the roots of the quadratic equation are and .

However, since this question involves lengths, we must ensure that the value of gives a positive base and height. The solution is not valid. For example, let’s substitute into the expression for the height

Height cm

We get

Height cm

Since a length cannot be negative, we discard Hence, the only solution is

It can be easily verified that this value of is correct as both the base and height of the triangle will be positive.