In this activity, we will be looking at more complex quadratic equations, which can be solved by factorising. For example:
2x² + x - 3 = 0
The method of factorising a quadratic expression of the form ax² + bx + c into two brackets is covered in another activity, and here we assume you have completed that activity, but if you haven't it would be best if you do that before continuing here.
So, the equation 2x² + x - 3 = 0 can be factorised into two brackets as (x - 1)(2x + 3) = 0. Now, remember that each set of brackets stand for a number and these are multiplied together to give the answer zero. So, if I tell you that I have multiplied together two numbers and the answer is zero can you tell me anything about my two numbers? Well, you should be able to tell me that at least one of my numbers is zero itself. Going back to my equation, that means that either (x - 1) = 0 or (2x + 3) = 0 (they cannot both be zero). But if x - 1 = 0 then x = 1 is a solution. Alternatively, if 2x + 3 = 0 then x = -3/2 (-1.5) is another solution. So there are two possible solutions to this quadratic equation, x = 1 and x = -1.5. it is a feature of quadratic equations that they nearly always have two solutions.
Here's another example, this time we'll just include the lines of working you would need to show.
6n² - 5n - 4 = 0
(2n + 1)(3n - 4) = 0
Either 2n + 1 = 0 or 3n - 4 = 0
So n = -1/2 or n = 4/3
Finally, we'll look at one more example.
4z(z - 1) = 3
Now, this equation looks like it has already been factorised. However, you will notice the right-hand side is 3, not 0. Quadratic equations can only be solved if the right-hand side is zero. So we need to rearrange the equation first and this means we must expand the brackets.
4z² - 4z = 3
Now subtract 3 from both sides.
4z² - 4z -3 = 0
Now it is in the correct form for us to factorise into two brackets and then solve.
4z² - 4z -3 = 0
(2z - 3)(2z + 1) = 0
Either 2z - 3 = 0 or 2z + 1 = 0
So z = 3/2 or z = -1/2
Ready to try some for yourself now? Have a go at the questions.