__About Me:__

I'm currently a first year undergraduate at University of York, studying Natural Sciences specializing in Mathematics, Chemistry and Physics. I have a real passion and drive when it comes to all things science and maths.

I'm very friendly, understanding and a good listener. I mentored at my previous school, helping students from age 14-17 and helped coach at my local swimming club, gaining great experience.

__Tutorials:__

My tutorials are all about the student. You will choose what to cover. I first will focus on the fundamental concepts of the particular module, until you are able to explain it yourself, then work on exam technique by using past exam questions. I place a big emphasis on clear concise workings as I believe this helps your understanding, as well as, improve exam technique.

I'm eager to help with GCSE mathematics and physics. I also have a very good understanding of A-level physics and mathematics, especially core modules, mechanics, further pure and statistics. I look forward to helping you over come scientific concepts and hopefully ultimately increase your knowledge and grade.

__What next?__

Either send me a direct ‘WebMail’ or send me a ‘Meet The Tutor Session’ along with the subject enquiry, specific modules that you have problems with and the exam board.

__About Me:__

I'm currently a first year undergraduate at University of York, studying Natural Sciences specializing in Mathematics, Chemistry and Physics. I have a real passion and drive when it comes to all things science and maths.

I'm very friendly, understanding and a good listener. I mentored at my previous school, helping students from age 14-17 and helped coach at my local swimming club, gaining great experience.

__Tutorials:__

My tutorials are all about the student. You will choose what to cover. I first will focus on the fundamental concepts of the particular module, until you are able to explain it yourself, then work on exam technique by using past exam questions. I place a big emphasis on clear concise workings as I believe this helps your understanding, as well as, improve exam technique.

I'm eager to help with GCSE mathematics and physics. I also have a very good understanding of A-level physics and mathematics, especially core modules, mechanics, further pure and statistics. I look forward to helping you over come scientific concepts and hopefully ultimately increase your knowledge and grade.

__What next?__

Either send me a direct ‘WebMail’ or send me a ‘Meet The Tutor Session’ along with the subject enquiry, specific modules that you have problems with and the exam board.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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__Finding Roots of Quadratic Equations__

__What is a root to an equation?__

A root to an equation is a set of value(s) that satisfy the equation and when shown graphically they are the x values at which the function intercepts the x-axis.

__The general form of a quadratic equation is:__

ax^{2} + bx + c = 0 where a, b and c are real coefficients

and before attempting to solve any quadratic function, you should always aim to get it into this form first.

If it is not in the correct form it can be converted by adding and subtracting each side functions of x in the initial form, for example:

x^{2 }= 2x – 12 (Subtract both sides by 2x)

x^{2 }– 2x = -12 (Add 12 to both sides)

x^{2 }– 2x + 12 = 0

As you can now see the previous equation is now in the standard form ax^{2} + bx + c = 0, where a = 1, b = 2 and c = 12.

__Finding the Root(s) of Quadratic Equations__

The first way to solve a quadratic equation is by __factorising__ it, an example is:

x^{2 }+ 7x + 12 = 0 --> (x + 3)(x + 4) = 0

the root to the equation is then given by the negative coefficient of the real number inside the bracket, hence is -3 and -4. A sketch of this graph would consist of a U shape intercepting the x-axis at -3 and -4.

The second way to solve a quadratic equation is to __complete the square__, an example is:

x^{2 }- 10x + 25 = 0 --> (x-5)^{2} = 9

The root to this equation is then worked out by square rooting each side and adding 5 to both sides, giving 8 and -2.

The final way to solve them by a __quadratic formula__, which is:

x = (-b +/- sqr(b^{2} – 4ac))/2a

The quadratic formula contains the function b^{2} – 4ac, this is called the discriminant and a, b and c are the coefficients of the equation when in the standard form. The value of the discriminant can show how many roots are present for a particular equation:

b^{2} – 4ac > 0 2 real roots

b^{2} – 4ac = 0 1 real root

b^{2} – 4ac < 0 2 imaginary roots (Complex conjugates)

__Example 1__

x^{2} + 6x + 3 = 0 a=1, b=6 and c=3

b^{2} – 4ac = 36 – 12 = 24

hence x = (-6 +/- 2sqr6)/2 = -3 +/- sqr6

so the two roots are -3 + sqr6 and -3 – sqr6

__Example 2__

x^{2} + 2x + 1 = 0 a=1, b=2 and c=1

b^{2} – 4ac = 0

hence x = -2/2 = -1

__Example 3__

x^{2} + 8x + 25 = 0 a=1, b=8 and c=25

b^{2} – 4ac = 64 – 100 = -36

The discriminant is less than 0, which shows that 2 complex conjugate roots are the solutions to the equation.

Since we can not find the square root of a negative number, we instead denote the term i, this represents the square root of -1 and also shows that i^{2} = -1. This now allows the solution to be found:

x = (-8 +/- 6i)/2 = -4 +/- 3i

hence the solutions are -4 + 3i and -4 – 3i which are complex conjugates

__Finding Roots of Quadratic Equations__

__What is a root to an equation?__

A root to an equation is a set of value(s) that satisfy the equation and when shown graphically they are the x values at which the function intercepts the x-axis.

__The general form of a quadratic equation is:__

ax^{2} + bx + c = 0 where a, b and c are real coefficients

and before attempting to solve any quadratic function, you should always aim to get it into this form first.

If it is not in the correct form it can be converted by adding and subtracting each side functions of x in the initial form, for example:

x^{2 }= 2x – 12 (Subtract both sides by 2x)

x^{2 }– 2x = -12 (Add 12 to both sides)

x^{2 }– 2x + 12 = 0

As you can now see the previous equation is now in the standard form ax^{2} + bx + c = 0, where a = 1, b = 2 and c = 12.

__Finding the Root(s) of Quadratic Equations__

The first way to solve a quadratic equation is by __factorising__ it, an example is:

x^{2 }+ 7x + 12 = 0 --> (x + 3)(x + 4) = 0

the root to the equation is then given by the negative coefficient of the real number inside the bracket, hence is -3 and -4. A sketch of this graph would consist of a U shape intercepting the x-axis at -3 and -4.

The second way to solve a quadratic equation is to __complete the square__, an example is:

x^{2 }- 10x + 25 = 0 --> (x-5)^{2} = 9

The root to this equation is then worked out by square rooting each side and adding 5 to both sides, giving 8 and -2.

The final way to solve them by a __quadratic formula__, which is:

x = (-b +/- sqr(b^{2} – 4ac))/2a

The quadratic formula contains the function b^{2} – 4ac, this is called the discriminant and a, b and c are the coefficients of the equation when in the standard form. The value of the discriminant can show how many roots are present for a particular equation:

b^{2} – 4ac > 0 2 real roots

b^{2} – 4ac = 0 1 real root

b^{2} – 4ac < 0 2 imaginary roots (Complex conjugates)

__Example 1__

x^{2} + 6x + 3 = 0 a=1, b=6 and c=3

b^{2} – 4ac = 36 – 12 = 24

hence x = (-6 +/- 2sqr6)/2 = -3 +/- sqr6

so the two roots are -3 + sqr6 and -3 – sqr6

__Example 2__

x^{2} + 2x + 1 = 0 a=1, b=2 and c=1

b^{2} – 4ac = 0

hence x = -2/2 = -1

__Example 3__

x^{2} + 8x + 25 = 0 a=1, b=8 and c=25

b^{2} – 4ac = 64 – 100 = -36

The discriminant is less than 0, which shows that 2 complex conjugate roots are the solutions to the equation.

Since we can not find the square root of a negative number, we instead denote the term i, this represents the square root of -1 and also shows that i^{2} = -1. This now allows the solution to be found:

x = (-8 +/- 6i)/2 = -4 +/- 3i

hence the solutions are -4 + 3i and -4 – 3i which are complex conjugates