## What you need to know

Quadratic equations are equations that include a squared term, usually an $x^2$. Typically, they are presented like

$x^2+7x+6=0$

If you are on the foundation course, any quadratic equation you’re expected to solve will always appear like this, with all terms on one side and a zero on the other. If you are on the higher course, you may have to do some rearranging in order to get all the terms on one side.

In any case, we wish to solve this equation, and in this topic,  we’re going to do so by using factorisation. If you’re not sure how to factorise a quadratic, click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/factorising-quadratics-gcse-maths-revision-worksheets/). To do see how this is done, let’s go right into an example.

Example: Use factorisation to solve $x^2+7x+6=0$.

So, we must factorise this quadratic. Observing that $6\times1=6$ and $6+1=7$, we get that the quadratic factorises to

$x^2+7x+6=(x+1)(x+6)$

Therefore, we can rewrite the quadratic equation in the question to be

$(x+1)(x+6)=0$

The question is, how will this help us solve it? Well, firstly recognise that the two brackets, $(x+1)$ and $(x+6)$ are being multiplied together (this is one of those things that we use and take for granted without really thinking about it, just like how when we write $7x$ we’re actually saying $7\times x$).

Then, since the right-hand side of the equation is zero, the result of multiplying the two brackets together on the left-hand side must be zero. Furthermore, if this is the case, then at least one of the brackets must be equal to zero. Think about it – if you multiply two numbers together and get zero as the answer, one of them must be zero. We’re just applying that rule to algebra. So, in summary,

$\text{if }(x+1)(x+6)=0,\,\text{ then either }\,(x+1)=0\,\text{ or }\,(x+6)=0.$

This now gives us two very simple equations to solve:

$x+1=0\,\text{ and }\,x+6=0$.

To solve the first equation, subtract 1 from both sides to get $x=-1$. To solve the second equation, subtract 6 from both sides to get $x=-6$. Thus, the two solutions to the equation $x^2+7x+6=0$ are $x=-1$ and $x=-6$.

You may have realised that the two solutions are just the numbers in the factorisation, $(x+1)(x+6)$, with opposite signs. This continues to be true, and once you get used to it you can skip the last step entirely and get the solutions quickly.

The solutions to a quadratic equation are also called roots, because they correspond to where a quadratic graph crosses the $x$-axis. In this example, we’ll see why that’s the case.

Example: Use factorisation to find the roots of $y=x^2+x-2$. Sketch the graph.

This question doesn’t actually give us a quadratic equation, but we’re going to set one up from the quadratic given and solve it. We want to find the solutions to

$x^2+x-2=0$.

Observing that $2\times(-1)=-10$ and $2+(-1)=1$, we can factorise this quadratic to be $(x+2)(x-1)$, and thus the equation becomes

$(x+2)(x-1)=0$

One of these brackets must be zero. Either $x+2=0$, in which case $x=-2$, or $x-1=0$, in which case $x=1$.

So, how are these values related to where the graph crosses the $x$-axis? Well, the graph we will sketch will have equation $y=x^2+x-2$, and we found the solutions to $x^2+x-2=0$. Put these two equations together, and we see that we found the values of $x$ for when

$y=x^2+x-2=0$

In other words, when $y=0$. Given that the $x$-axis is precisely the line $y=0$, we now know that the solutions we found: $x=-2$ and $x=1$, are the points where $y=0$ and thus where the graph crosses the $x$-axis. We call these the roots of the quadratic.

Once you have found these values, and you know the general shape of a quadratic graph, that is enough to sketch it.

This picture on the right was done by a computer, but a “sketch” doesn’t have to be perfect, it just has to be the right shape and cross at the right points.

If you are on the higher course, you may have to find solutions by factorising quadratics such as

$2x^2-3x-9=0$

You should be able to factorise this to $(2x+3)(x-3)$ and thus write the equation as

$(2x+3)(x-3)=0$

At this point, the process is exactly the same. If the second bracket is zero: $x-3=0$, then we get $x=3$. If the first bracket is zero: $2x+3=0$, then we have to put a bit more effort in, but this is the only real difference. If we subtract 3 from both sides, and then divide both sides by 2, we get the solution to be

$x=-\dfrac{3}{2}$.

So, the 2 solutions to the equation are $x=3$ and $x=-\frac{3}{2}$.

## Solving Quadratics Through Factorising Questions

So, we want to factorise $p^2-3p-10$. Observing that $2\times(-5)=-10$ and $2+(-5)=-3$, we get that this quadratic factorises to

$p^2-3p-10=(p+2)(p-5)$

Therefore, we can rewrite the equation as

$(p+2)(p-5)=0$

So, for the left-hand side to be zero we require one of the brackets to be zero. If the first bracket is zero, then we get

$p+2=0,\,\text{ which admits the solution }\,p=-2$

If the second bracket is zero, then we get

$p-5=0,\,\text{ which admits the solution }\,p=5$

Therefore, the two solutions to this quadratic equation are $p=-2$ and $p=5$.

To find the roots of a quadratic, we must set it equal to zero and find the solutions of that equation:

$x^2-6x+8=0$

Now, we must factorise. Observing that $(-2)\times(-4)=8$ and $-2+(-4)=-6$, we get that this quadratic factorises to

$x^2-6x+8=(x-2)(x-4)$

Therefore, we can rewrite the equation as

$(x-2)(x-4)=0$

So, for the left-hand side to be zero we require one of the brackets to be zero. If the first bracket is zero, then we get

$x-2=0,\,\text{ which admits the solution }\,x=2$

If the second bracket is zero, then we get

$x-4=0,\,\text{ which admits the solution }\,x=4$

Therefore, the two roots of this quadratic equation are $x=2$ and $x=4$. Since these roots are precisely the points that the graph crosses the $x$-axis, we can now sketch this graph. Given that it is a positive quadratic and we know the general shape of it, the resulting graph should look like

So, we want to factorise $3k^2+17k-6$. Firstly, we will do: $3\times(-6)=-18$. Now, observing that $18\times(-1)=-18$ and $18+(-1)=17$, we will rewrite this quadratic as

$3k^2-k+18k-6$

Then, factorising the first two terms and last two terms separately, we get

$k(3k-1)+6(3k-1)$

Therefore, we get that this quadratic factorises to

$(3k-1)(k+6)$

So, we can rewrite the equation as

$(3k-1)(k+6)=0$

So, for the left-hand side to be zero we require one of the brackets to be zero. If the first bracket is zero, then we get

$3k-1=0$

If we add 1 to both sides and then divide by 3, we get the solution

$k=\dfrac{1}{3}$

If the second bracket is zero, then we get

$k+6=0,\,\text{ which admits the solution }\,k=-6$

Therefore, the two solutions to this quadratic equation are $k=\frac{1}{3}$ and $k=-6$.