Solving Quadratics Through Factorising Worksheets | MME

# Solving Quadratics Through Factorising Worksheets, Questions and Revision

Level 6 Level 7

## What you need to know

### Solving Quadratic Equations by Factorising

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.

If you’re not sure how to factorise a quadratic, revise factorising quadratics

### Take Note:

Once you have learnt how to factorise quadratics, solving them is actually the easier part of the question.

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. Therefore, at least one of the brackets must be equal to zero. So, in summary,

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

Note:The solutions to a quadratic equation are also called roots, because they correspond to where a quadratic graph crosses the $x$-axis.

### Example 1: Solving Quadratic Equations by Factorising

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

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

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

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

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

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 are $x=-1$ and $x=-6$.

Note: You may have realised that the two solutions are just the numbers in the factorisation, $(x+1)(x+6)$, with opposite signs.

### Example 2: Quadratic Equations and Graphs

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

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

So $x=-2$ or $x=1$.

So, how are these values related to where the graph crosses the $x$-axis?

Given that the $x$-axis is precisely the line $y=0$, we now know that the solutions 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, you can sketch the graph, see below.

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

### Example 3: Solving Quadratic Equations by Factorising

Find solutions by factorising

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

Using your knowledge of factorising quadratics you get $(2x+3)(x-3)$ 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 $x=-\frac{3}{2}$.

### Example Questions

The quadratic on the left hand side of the equation factorises so that,

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

Therefore, we can rewrite the equation as

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

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

$p=-2$ and $p=5$

The quadratic on the left hand side of the equation factorises so that,

$x^2-8x+15=(x-5)(x-3)$

Therefore, we can rewrite the equation as

$(x-5)(x-3)=0$

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

$x=5$ and $x=3$

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

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

So, for the left-hand side to be zero we require one of the brackets to be zero.

Therefore, the two roots of this quadratic equation are

$x=2$ and $x=4$

The quadratic on the left hand side of the equation factorises so that,

$2x^2+13x+15=(2x+3)(x+5)=0$

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

$x=-\dfrac{3}{2}$ and $x=-5$

Firstly, as $3\times(-6)=-18$, and 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

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

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

$k=-6$

Therefore, the two solutions to this quadratic equation are,

$k=\frac{1}{3}$ and $k=-6$

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