I'm a Physics (MPhys) student currently studying at Oxford. I think that, often, it's easier to learn from a student than from a teacher - a teacher might not remember the process of learning a new skill or concept, and so it seems natural and obvious to them why things are the way they are. On the other hand, a student will remember getting things wrong at first - a crucial and vital part of science! - and how they were able to remember and justify things to themselves.

A key part of gaining ability in science and maths is seeing how things fit together into a single framework that makes sense to the student. This makes new concepts easier to understand and remember, and allows you to check your own answers in exams - a luxury that not every student has!

I'm a Physics (MPhys) student currently studying at Oxford. I think that, often, it's easier to learn from a student than from a teacher - a teacher might not remember the process of learning a new skill or concept, and so it seems natural and obvious to them why things are the way they are. On the other hand, a student will remember getting things wrong at first - a crucial and vital part of science! - and how they were able to remember and justify things to themselves.

A key part of gaining ability in science and maths is seeing how things fit together into a single framework that makes sense to the student. This makes new concepts easier to understand and remember, and allows you to check your own answers in exams - a luxury that not every student has!

No DBS Check

There are several different methods for solving quadratic equations. The method you use depends on the equation in question.

Usually, the easiest method is by factorising the quadratic, so you want to check if you can do this first. Write the quadratic as

Ax^{2} + Bx + C = 0

Then consider the expression

(Ax + )(x + ) = 0

Where spaces have been left for real numbers. This will clearly multiply to give you the term in x^{2} so now the challenge is to find a pair of numbers that multiply together to give C and will multiply with the term in x to give B.

It might help to factorise C - which numbers can multiply together to give you C? Then trialling these numbers and seeing what coefficient they give you for x will enable you to find the factorisation, if it exists.

For example:

2x^{2} + 7x + 6 = 0

Write as (2x + ) (x + )

Note that numbers that will multiply to give us 6 are (6, 1), and (2, 3). Trying 6 and 1 quickly shows these numbers will not work, and we can see that:

(2x + 3 )(x + 2) = 0

when multiplied out gives the appropriate coefficient of 7 for x. The solutions are then the values of x that can make the contents of the brackets equal zero:

x = - 1.5 and x = -2.

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The quadratic formula works regardless of whether or not the quadratic can be factorised. However, it is algebraically more complicated, so factorising is preferable if you can do it quickly. The quadratic formula gives the solutions as:

x = (- B + (B^{2} - 4AC)^{0.5 })/ 2A

x = (- B - (B^{2} - 4AC)^{0.5 })/ 2A

where A, B and C are the coefficients of a quadratic Ax^{2} + Bx + C = 0.

Note that if B^{2} - 4AC is less than zero, this involves taking the square root of a negative number, and so the quadratic will have no real solutions. However, complex solutions - mixtures of real and imaginary numbers - will exist.

There are several different methods for solving quadratic equations. The method you use depends on the equation in question.

Usually, the easiest method is by factorising the quadratic, so you want to check if you can do this first. Write the quadratic as

Ax^{2} + Bx + C = 0

Then consider the expression

(Ax + )(x + ) = 0

Where spaces have been left for real numbers. This will clearly multiply to give you the term in x^{2} so now the challenge is to find a pair of numbers that multiply together to give C and will multiply with the term in x to give B.

It might help to factorise C - which numbers can multiply together to give you C? Then trialling these numbers and seeing what coefficient they give you for x will enable you to find the factorisation, if it exists.

For example:

2x^{2} + 7x + 6 = 0

Write as (2x + ) (x + )

Note that numbers that will multiply to give us 6 are (6, 1), and (2, 3). Trying 6 and 1 quickly shows these numbers will not work, and we can see that:

(2x + 3 )(x + 2) = 0

when multiplied out gives the appropriate coefficient of 7 for x. The solutions are then the values of x that can make the contents of the brackets equal zero:

x = - 1.5 and x = -2.

------------------------------------------------------------------------

The quadratic formula works regardless of whether or not the quadratic can be factorised. However, it is algebraically more complicated, so factorising is preferable if you can do it quickly. The quadratic formula gives the solutions as:

x = (- B + (B^{2} - 4AC)^{0.5 })/ 2A

x = (- B - (B^{2} - 4AC)^{0.5 })/ 2A

where A, B and C are the coefficients of a quadratic Ax^{2} + Bx + C = 0.

Note that if B^{2} - 4AC is less than zero, this involves taking the square root of a negative number, and so the quadratic will have no real solutions. However, complex solutions - mixtures of real and imaginary numbers - will exist.