About Me:
I am a mathematics student at Warwick. I pride myself on having a deep and thorough understanding of mathematical concepts from A level to below. I feel I have encountered the most common classic mistakes mathematics students make as I have made them myself. Throughout my mathematics career I have made sure that I rectify any small misunderstanding I have and I feel this puts me in a great position to teach others about the subject I love.
I have privately tutored students at mathematics GCSE and A level and I have also volunteered for a year in primary school mathematics lesson as a classroom assistant. I believe everyone can be great at mathematics and that any self doubt is usually due to small fixable mistakes in the understanding of basic concepts.
The Sessions:
I will focus primarily on any particular questions you may have for me. If we get through these questions I like to work on exam style questions as this is the best way to improve your grade. If it becomes clear that there is a fundamental gap n your knowledge on a certain topic then we will take a step back and slowly go through the theory of that topic.
For this style of tutoring the exam board you are on is very important so the sooner I could get this information the better.
Oxford, MAT, STEP:
I have personally had interviews for Oxford for a mathematics degree and will try my best to advise you on interview technique and share my experience. I have also achieved a 2 in STEP 1 and would be willing to look over any questions you might have about it, although I must stress, unlike A Level, my knowledge on STEP is far from comprehensive. I also passed the MAT and would be wiling to advise on that front.
Questions:
For any questions send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website).
About Me:
I am a mathematics student at Warwick. I pride myself on having a deep and thorough understanding of mathematical concepts from A level to below. I feel I have encountered the most common classic mistakes mathematics students make as I have made them myself. Throughout my mathematics career I have made sure that I rectify any small misunderstanding I have and I feel this puts me in a great position to teach others about the subject I love.
I have privately tutored students at mathematics GCSE and A level and I have also volunteered for a year in primary school mathematics lesson as a classroom assistant. I believe everyone can be great at mathematics and that any self doubt is usually due to small fixable mistakes in the understanding of basic concepts.
The Sessions:
I will focus primarily on any particular questions you may have for me. If we get through these questions I like to work on exam style questions as this is the best way to improve your grade. If it becomes clear that there is a fundamental gap n your knowledge on a certain topic then we will take a step back and slowly go through the theory of that topic.
For this style of tutoring the exam board you are on is very important so the sooner I could get this information the better.
Oxford, MAT, STEP:
I have personally had interviews for Oxford for a mathematics degree and will try my best to advise you on interview technique and share my experience. I have also achieved a 2 in STEP 1 and would be willing to look over any questions you might have about it, although I must stress, unlike A Level, my knowledge on STEP is far from comprehensive. I also passed the MAT and would be wiling to advise on that front.
Questions:
For any questions send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website).

No DBS Check

4.7from 4 customer reviews

Andrea (Parent)

October 3 2016

Kirsty (Parent)

September 27 2016

Kirsty (Parent)

September 23 2016

Antonella (Parent)

July 31 2016

Where the graph intersects the x-axis, x^{2}+2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x^{2}+2kx+5 = 0 not have a solution?

This is now a simpler problem (roots of a quadratic equation). We can apply the common method of considering the discriminant of x^{2}+2kx+5. Using standard quadratic formula notation where in this case a=1, b =2k and c=5 we evaluate the discriminant : b^{2}-4ac= (2k)^{2}-4*1*5 = 4k^{2} -20.

Now since the discriminant appears in a square root sign in the quadratic equation, if it is negative there can be no real solutions to the equation ( great this is what we want!).

Thus we want discriminant negative: 4k^{2} -20 <0. Divide both sides of the inequality by 4 so we have k^{2}-5<0.

Now this is where we must take great care, the following reasoning is a common ** MISTAKE**: rearragne the inequality so we have k

When dealing with inequalities involving powers such as we are here we must be extremely careful. the mistake in the reasoning above is when we say k < - sqrt(5), this is actually a form of the common mistake of not inverting the inequality when multiplying both sides of an equation by a negative. Instead when dealing with inequalities with powers it is always much wiser to sketch a graph of the situation.

k^{2} - 5 is the standard quadratic U shape (think y=x^{2}) shifted down by 5. Having sketched this out it is clear that this graph is less then 0 when it is inbetween it's two roots.

The roots of k^{2 }-5 are easy to find: k^{2 }-5 = 0 implies k^{2} = 5 implies k = sqrt(5) or k = -sqrt(5).

Comparing this with the graph we can now see that the discriminant is negative for - sqrt(5) < k < sqrt(5). Thus these are the values for which the graph y=x^{2}+2kx+5 does not intersect the x-axis.

Where the graph intersects the x-axis, x^{2}+2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x^{2}+2kx+5 = 0 not have a solution?

This is now a simpler problem (roots of a quadratic equation). We can apply the common method of considering the discriminant of x^{2}+2kx+5. Using standard quadratic formula notation where in this case a=1, b =2k and c=5 we evaluate the discriminant : b^{2}-4ac= (2k)^{2}-4*1*5 = 4k^{2} -20.

Now since the discriminant appears in a square root sign in the quadratic equation, if it is negative there can be no real solutions to the equation ( great this is what we want!).

Thus we want discriminant negative: 4k^{2} -20 <0. Divide both sides of the inequality by 4 so we have k^{2}-5<0.

Now this is where we must take great care, the following reasoning is a common ** MISTAKE**: rearragne the inequality so we have k

When dealing with inequalities involving powers such as we are here we must be extremely careful. the mistake in the reasoning above is when we say k < - sqrt(5), this is actually a form of the common mistake of not inverting the inequality when multiplying both sides of an equation by a negative. Instead when dealing with inequalities with powers it is always much wiser to sketch a graph of the situation.

k^{2} - 5 is the standard quadratic U shape (think y=x^{2}) shifted down by 5. Having sketched this out it is clear that this graph is less then 0 when it is inbetween it's two roots.

The roots of k^{2 }-5 are easy to find: k^{2 }-5 = 0 implies k^{2} = 5 implies k = sqrt(5) or k = -sqrt(5).

Comparing this with the graph we can now see that the discriminant is negative for - sqrt(5) < k < sqrt(5). Thus these are the values for which the graph y=x^{2}+2kx+5 does not intersect the x-axis.

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, and label the hypotenuse H.

Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We also know by pythagoras that A^{2 }+ O^{2} = H^{2} . We shall now combine these equations to get our result.

A/H = 1/3 implies H = 3A implies H^{2} = 9A^{2}. Substituting this result into our euation obtained by pythagoras we get: A^{2 }+ O^{2} = 9A^{2}. Rearranging: O^{2} = 8A^{2} implies O^{2}/A^{2} = 8 implies (O/A)^{2} = 8. Now we take the square root of both sides. Here we must take care, O and A are lengths and so are not negative, so we only consider the positive root: O/A = sqrt(8) = tan(x) and so we are done.

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, and label the hypotenuse H.

Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We also know by pythagoras that A^{2 }+ O^{2} = H^{2} . We shall now combine these equations to get our result.

A/H = 1/3 implies H = 3A implies H^{2} = 9A^{2}. Substituting this result into our euation obtained by pythagoras we get: A^{2 }+ O^{2} = 9A^{2}. Rearranging: O^{2} = 8A^{2} implies O^{2}/A^{2} = 8 implies (O/A)^{2} = 8. Now we take the square root of both sides. Here we must take care, O and A are lengths and so are not negative, so we only consider the positive root: O/A = sqrt(8) = tan(x) and so we are done.

Let us begin by simplifying the expression:

(n+3)^{2} - n^{2} = (n+3)(n+3) - n^{2}

= n^{2} + 6n + 9 - n^{2} (expanded brackets)

= 6n + 9 (collected like terms)

= 3(2n+3) (taken out a factor of 3)

Now we can consider this simpler equivalent expression.

3 is an odd number

2n is even thus 2n+3 is odd (even plus odd is always odd)

so we have an odd*odd which is always odd, thus never even and we are done.

Let us begin by simplifying the expression:

(n+3)^{2} - n^{2} = (n+3)(n+3) - n^{2}

= n^{2} + 6n + 9 - n^{2} (expanded brackets)

= 6n + 9 (collected like terms)

= 3(2n+3) (taken out a factor of 3)

Now we can consider this simpler equivalent expression.

3 is an odd number

2n is even thus 2n+3 is odd (even plus odd is always odd)

so we have an odd*odd which is always odd, thus never even and we are done.