If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.
This week, we’re exploring:
- Solving Quadratics by Factorising
- Reflection
- Multiplying Mixed Numbers
- Circle Theorems
- Angles in Parallel Lines
Question 1: Solving Quadratics by Factorising
Factorising is one way to solve quadratic equations. It doesn’t work for all quadratics but, if the question is on a non-calculator paper, the chances are you can solve the quadratic by factorising it.
Solve \(6x^2 – 19x + 15 = 0\).
First, we factorise the left-hand side. You may want to look back at week 5 if you think you need to revisit this.
To factorise, we look for a pair of numbers that multiply together to give 𝑎𝑐 = 90 and add together to give -19. The only pair of numbers that fits both requirements is -9 and -10. We use this to split our middle term and then factorise in pairs:
\(\begin{aligned}6x^2-9x-10x+15 &= 3x(2x-3)-5(2x-3)\\ &= (3x-5)(2x-3) \end{aligned}\)
So, \((3x-5)(2x-3)=0\). For two numbers to multiply together to give 0, one of them must be zero.
\(3x-5=0\)
\(x=\frac{5}{3}\)
or
\(2x-3=0\)
\(x=\frac{3}{2}\)
So, the two solutions to \(6x^2 – 19x + 15 = 0\) are \(x=\frac{5}{3}\) and \(x=\frac{3}{2}\) .
Question 2: Reflection
When reflecting a shape, you will be given a straight line to reflect a shape in. Usually, you will need to draw this line in first.
Reflect the shape below in the line with equation 𝑦 = 𝑥.
First, we need to draw the line with equation 𝑦 = 𝑥. This will go through the points (0, 0), (1, 1), (2, 2) and so on.
Now, we think about how the reflection will affect each vertex (corner) of the shape. If we consider the perpendicular distance to the line, the reflected vertex will be the same distance on the other side of the line.
Joining these three vertices together gives us our final shape:
Describe the single transformation that maps shape A onto shape B.
The key here is to find the straight line that is the same distance from shape A and from shape B.
This is the line with equation 𝑥 = 5.
Question 3: Multiplying Mixed Numbers
When multiplying fractions, we simply multiply the numerators (the top numbers) and then multiply the denominators (the bottom numbers). Then, if we need to, we simplify our answer. With mixed numbers, we need to change them to improper fractions first.
Let’s say we want to calculate \( 2\frac{1}{7}\times 1\frac{5}{6}\).
\( \begin{aligned} 2\frac{1}{7}\times 1\frac{5}{6} &= \frac{15}{7}\times \frac{11}{6} \\ &= \frac{15 \times 11}{7 \times 6} \\ &= \frac{165}{42} \end{aligned}\)
We need to see if we can simplify our answer by checking for any common factors of 165 and 42. In this case, they are both divisible by 3 so we divide both the numerator and denominator by 3:
\( \begin{aligned} \frac{165}{42} = \frac{55}{14}\end{aligned}\)
55 and 14 have no common factors so this fraction is in its simplest form. Now, we just put it back as a mixed number.
\( \begin{aligned} \frac{55}{14} = 3\frac{13}{14}\end{aligned}\)
Question 4: Circle Theorems
These are the two circle theorems that are relevant this week.
Given that O is the centre of the circle and PQ is a tangent to the circle, find the size of the angle marked 𝑥.
Alternate segment theorem means that angle BCQ = angle BAC so angle BCQ = 28°.
OC is a radius and CQ is a tangent, so angle OCQ = 90°.
Therefore:
\( \begin{aligned} x &= 90-28 \\ &= 62° \end{aligned} \)
Question 5: Angles in Parallel Lines
There are three rules you need to know for angles in parallel lines:
You might also need to use these rules – they are often used in questions that include angles in parallel lines.
Let’s find the labelled missing angles in the shape below.
The best way to start is to write down any angles you can see. For example, you might spot that 83° and 
= 97° Angles on a straight line add up to 180°.
Write as much detail as you can in the reason – most exam boards have keywords they are looking for, so it is worth looking at mark schemes to see what these are.
Next, we might spot that 
Finally, you could spot that 
You could also have spotted that
There is always more than one way to answer a question like this. The reasons are worth a large proportion of the marks so if you can see more than one solution, you should use the one that you can explain better.
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