Hi i'm Alex. I am currently a first year Maths student at Warwick University, and can tutor you in Maths, Further Maths and Physics at both GCSE and A-Level. I have had plenty of teaching experience before so I can hopefully help you with any questions you have.

Hi i'm Alex. I am currently a first year Maths student at Warwick University, and can tutor you in Maths, Further Maths and Physics at both GCSE and A-Level. I have had plenty of teaching experience before so I can hopefully help you with any questions you have.

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'Completing the square' is quite a tricky concept to some people, and I honestly didnt grasp it the first time i was taught it. But once explained thoroughly, it becomes easier to use.

'Completing the square' is a method for solving quadratic equations, when an equation cannot be easily factorised. In fact, the quadratic formula you will see in formulae books, is proven by 'completing the square' of the following equation: ax^{2} + bx + c = 0.

I am going to explain the 'completing the square' method by using the following example.

I will use the equation x^{2} - 6x + 5 =0.

Those of you who are quite eagle eyed will notice that you can easily factorise this equation and see that the solutions are x=1 and x=5, but i want to show you how to find these solutions by 'completing the square'.

The first step involves putting the x into a bracket with a squared on the outside. To do this you need to look at the first two terms: x^{2} - 6x.

After the first step the equation should look like this: (x-3)^{2} - 9 + 5 = 0. I will explain why you do this. When you look at x^{2} -6x you need to ask yourself the follwing question: what expression in x can i square to get these two terms?

By asking yourself this question you might notice that if you are going to square an expression, then the number within the expression should be half the number of x's in your original question. This is how x^{2} - 6x becomes (x-3)^{2}.

In reality, the x^{2} - 6x becomes (x-3)^{2} - 9. The reason this happens lies in the expansion of (x+3)^{2}. When we expand the bracket we get x^{2} - 6x + 9, which is not what we want as it is 9 more than the expression we want. This is why the -9 appears to fix this problem.

Now we have (x - 3)^{2} - 9 + 5 = 0. We can condense this down to (x - 3)^{2} - 4 =0. The next step is to isolate the squared bracket; this means writing the equation as (x - 3)^{2} = 4.

As you can see, the equation is looking a lot nicer know, and the next step is to square root both sides of the equation. This will make two equations as the square root of 4 is both 2 and -2.

We know have x - 3 = 2 and x - 3 = -2. Adding 3 to both sides of both equations gives the required results of x = 5 and x = 1.

And there you have it, completing the square is best done when the coefficient of x^{2} is 1, but if you have a different coefficient, just divide all the numbers in the equation by the coefficient and then complete the square.

'Completing the square' is quite a tricky concept to some people, and I honestly didnt grasp it the first time i was taught it. But once explained thoroughly, it becomes easier to use.

'Completing the square' is a method for solving quadratic equations, when an equation cannot be easily factorised. In fact, the quadratic formula you will see in formulae books, is proven by 'completing the square' of the following equation: ax^{2} + bx + c = 0.

I am going to explain the 'completing the square' method by using the following example.

I will use the equation x^{2} - 6x + 5 =0.

Those of you who are quite eagle eyed will notice that you can easily factorise this equation and see that the solutions are x=1 and x=5, but i want to show you how to find these solutions by 'completing the square'.

The first step involves putting the x into a bracket with a squared on the outside. To do this you need to look at the first two terms: x^{2} - 6x.

After the first step the equation should look like this: (x-3)^{2} - 9 + 5 = 0. I will explain why you do this. When you look at x^{2} -6x you need to ask yourself the follwing question: what expression in x can i square to get these two terms?

By asking yourself this question you might notice that if you are going to square an expression, then the number within the expression should be half the number of x's in your original question. This is how x^{2} - 6x becomes (x-3)^{2}.

In reality, the x^{2} - 6x becomes (x-3)^{2} - 9. The reason this happens lies in the expansion of (x+3)^{2}. When we expand the bracket we get x^{2} - 6x + 9, which is not what we want as it is 9 more than the expression we want. This is why the -9 appears to fix this problem.

Now we have (x - 3)^{2} - 9 + 5 = 0. We can condense this down to (x - 3)^{2} - 4 =0. The next step is to isolate the squared bracket; this means writing the equation as (x - 3)^{2} = 4.

As you can see, the equation is looking a lot nicer know, and the next step is to square root both sides of the equation. This will make two equations as the square root of 4 is both 2 and -2.

We know have x - 3 = 2 and x - 3 = -2. Adding 3 to both sides of both equations gives the required results of x = 5 and x = 1.

And there you have it, completing the square is best done when the coefficient of x^{2} is 1, but if you have a different coefficient, just divide all the numbers in the equation by the coefficient and then complete the square.

When investigating graphs, you will often be asked to pick out features of the graph; stationary points being the most popular. You will need to know that a stationary point on f(x) can be found by solving the following equation: f'(x)=0.

Once you have found the stationary points, you will need to find the second derivative of the graph, also known as f''(x). By finding the values of f''(x) at the x-coordinates where stationary points exist, you can categorise the stationary points.

If f''(x) > 0, then the stationary point is a minimum point.

If f''(x) < 0, then the stationary point is a maximum point.

If f''(x) = 0, then the stationary point is a point of inflection.

When investigating graphs, you will often be asked to pick out features of the graph; stationary points being the most popular. You will need to know that a stationary point on f(x) can be found by solving the following equation: f'(x)=0.

Once you have found the stationary points, you will need to find the second derivative of the graph, also known as f''(x). By finding the values of f''(x) at the x-coordinates where stationary points exist, you can categorise the stationary points.

If f''(x) > 0, then the stationary point is a minimum point.

If f''(x) < 0, then the stationary point is a maximum point.

If f''(x) = 0, then the stationary point is a point of inflection.

People doing mathematics at A-Level will be familiar with the concept of using x and y coordinates in the cartesian coordinate system. But introducing the idea of the polar coordinate system can be confusing to some.

In the polar coordinate system, instead of using a square grid to plot points on, you use a circular grid. The centre of the grid is known as the pole.

Also, instead of the two axes we are familiar with, there is just one in the polar system. This axis is known as the initial line, and bears similarity to the positive x-axis in the cartesian system.

Know that we have the polar grid explained, we need to know how to plot coordinates upon it.

A polar coordinate can be plotted by knowing its magnitude and angle. The magnitude of a polar point is its distance from the pole and is represented by the letter ** r**.

The angle of a polar point is the angle between the initial line and a figurative line joining the point to the pole, and is measured anticlockwise from the intial line, and is measured in radians. The angle is denoted by **θ**.

So in summary, polar coordinates use a circular grid and can be shown in the form **(r, θ).**

People doing mathematics at A-Level will be familiar with the concept of using x and y coordinates in the cartesian coordinate system. But introducing the idea of the polar coordinate system can be confusing to some.

In the polar coordinate system, instead of using a square grid to plot points on, you use a circular grid. The centre of the grid is known as the pole.

Also, instead of the two axes we are familiar with, there is just one in the polar system. This axis is known as the initial line, and bears similarity to the positive x-axis in the cartesian system.

Know that we have the polar grid explained, we need to know how to plot coordinates upon it.

A polar coordinate can be plotted by knowing its magnitude and angle. The magnitude of a polar point is its distance from the pole and is represented by the letter ** r**.

The angle of a polar point is the angle between the initial line and a figurative line joining the point to the pole, and is measured anticlockwise from the intial line, and is measured in radians. The angle is denoted by **θ**.

So in summary, polar coordinates use a circular grid and can be shown in the form **(r, θ).**