# Daily Maths Revision – Week 15 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

If it’s a formula you need, here are the ones that are relevant this week.

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The Cosine Rule
$$a^2=b^2+c^2-2bc\cos A$$

This week, we’re exploring:

### Question 1: Solving Quadratic Equations Using the Formula

When we can’t solve a quadratic equation using factorising, we can use the formula. For the formula to work, we need to have the equation in the form $$ax^2+bc+c=0$$.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Find the exact solutions to $$3x^2-4x=2$$.

First, we need to rearrange the formula so it’s equal to 0. We do this by subtracting 2 from both sides.

\begin{aligned}3x^2-4x&=2 \\ 3x^2-4x-2&=0\end{aligned}

Now we can identify the values of $$a$$, $$b$$ and $$c$$. In this case, $$a=3$$, $$b=\text{-}4$$ and $$c=\text{-}2$$. Remember to include the sign with the coefficient.

\begin{aligned}x&=\frac{\text{-}b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{\text{-}(\text{-}4)\pm\sqrt{(\text{-}4)^2-4\times3\times(\text{-}2)}}{2\times3}\end{aligned}

Now we can put this into our calculator. We will need to do this twice: once where we add the square root and once where we subtract the square root.

\begin{aligned}x&=\frac{\text{-}(\text{-}4)+\sqrt{16+24}}{6} \\ &=\frac{2+\sqrt{10}}{3}\end{aligned}

\begin{aligned}x&=\frac{\text{-}(\text{-}4)-\sqrt{16+24}}{6} \\ &=\frac{2-\sqrt{10}}{3}\end{aligned}

### Question 2: Vector Arithmetic

Vector addition can be done numerically or by using diagrams. Remember, the top part of the vector gives information about how far to the right a vector goes. The lower part of the vector is how far up the vector goes.

Evaluate: $$\boldsymbol{2\begin{pmatrix} 3\\\text{-}1 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix}}$$.

To do this question numerically, we simply multiply the first vector by two, then add the second vector. We treat the top number and bottom numbers completely separately.

\begin{aligned} 2\begin{pmatrix} 3\\\text{-}1 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix} &= \begin{pmatrix} 6\\\text{-}2 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix} \\ &=\begin{pmatrix} 6-7\\\text{-}2-1 \end{pmatrix} \\ &= \boldsymbol{\begin{pmatrix} \text{-}1\\\text{-}3 \end{pmatrix}} \end{aligned}

This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down:

We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down.

The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour.

This is the vector $$\boldsymbol{\begin{pmatrix} \text{-}1\\\text{-}3 \end{pmatrix}}$$

### Question 3: Rotation

When rotating a shape, you will be given a centre of rotation, an angle and a direction. The only exception to this is if you are rotating through 180°; in this case, the direction is irrelevant.

You should always ask for tracing paper when rotating – it makes the task much easier. If working at home, baking paper makes a good alternative.

Let’s rotate this shape 90° anticlockwise about the point (2, 4).

First, we need to identify the centre of rotation – the point (2, 4). Then, we place our tracing paper over the image and draw both the shape and the point.

Now, use a pencil to hold the point in the same place and rotate the shape through 90° anticlockwise.

Finally, remove the tracing paper and draw the shape in the new position. When you’ve drawn the shape, you should check the answer by putting the tracing paper back.

If asked to describe a transformation, you can identify it as a rotation if the shape has turned. In some cases, however, this could also be a reflection so make sure you consider both. You then need to specify that it’s a rotation and give the angle and direction.

Describe the single transformation that maps shape A onto shape B.

The easiest way to find the centre of rotation is by trial and improvement. We can draw shape A and try a few different centre of rotations until we find the one that maps to B.

Let’s start by trying the point (4, 3). Draw shape A on your tracing paper and rotate it until your shape has the same orientation as shape B.

As you can see, the shape is too far to the left and too far down. Let’s try a point a little further up – say (4, 4).

Now the orange shape overlays shape B so this is the correct centre of rotation. Now we just need to write down what we did.

A rotation, centre (4, 4), 90° clockwise.

### Question 4: The Cosine Rule

By this point, you are probably familiar with the trigonometric ratios and how they work in right-angled triangles. In other words, you know how to use SOHCAHTOA.

The cosine rule extends the use of trigonometry to non-right-angled triangles. In 2023, you will be given the cosine rule: $$a^2 = b^2 + c^2 – 2bc \cos A$$

For the following triangle, let’s calculate the length of the missing side labelled $$x$$, correct to 2 decimal places.

We start by labelling each of the sides. We want the given angle to be A. Then, each lower-case letter is opposite the corresponding upper-case letter:

Now, we can substitute the values into the cosine rule:

\begin{aligned} a^2&=b^2+c^2-2bc\cos A \\ a^2&=11^2+13^2-2\times11\times13\times\cos(50) \\ &= 106.16… \\ a &= \sqrt{106.16…} \\ &= 10.30\text{cm} (2\text{d.p.})\end{aligned}

Let’s calculate the size of the angle BAC. Give your answer correct to 2 decimal places.

This time, the triangle is already labelled so we can substitute the values straight into the formula. There are two options here: either you can learn the second version of the formula and use this, or you can substitute the values into the original formula and then rearrange.

We have used the second version of the formula: $$\cos A =\frac{b^2+c^2-a^2}{2bc}$$

\begin{aligned}\cos A &=\frac{12^2+8^2-14^2}{2\times12\times8} \\&=0.0625 \\ A &= \cos^{-1}(0.0625) \\ &= 86.42° (2\text{d.p.})\end{aligned}

### Question 5: Volume of a Prism

A prism is any 3D shape with straight edges that has a consistent cross-section. To find the volume of a prism we simply find the area of the cross-section and multiply it by the length of the prism.

Let’s find the volume of the triangular prism below.

First, we need to find the area of the cross-section. In this case, that’s the triangle. In some cases, you might be given this area – if this is the case you can skip this step.

The height of this triangle is 2cm and the base is 6cm. So, using the formula for the area of a triangle:

area = $$\frac{1}{2} × base × height$$
= $$\frac{1}{2} × 2 × 6$$
= $$6$$cm2

Now, we simply multiply this area by the length of the prism; 10cm.

Volume = $$6\times10$$
= $$60$$cm3

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