How can you solve a quadratic equation?

There are 3 main ways of solving a quadratic equation x^2 + bx + c = 0 (we'll suppose the coefficient of x^2 is 1 for now). The first is factorisation, where if you can spot that you can factorise the expression into two linear factors, say x^2 + 5x + 6 = 0 factorises as (x+2)(x+3) = 0, then for the product to be 0 at least one of the linear factors x+2 and x+3 must be 0, so x is -2 or -3.

Another way is completing the square, where we rewrite the expression in the form (x + k)^2 - l for some numbers k and l. If the quadratic is in the form x^2 + bx + c as above, then we will always have k = b/2, and b can be worked out from this. For instance, for x^2 - 6x + 5, we will have k = -6/2 = -3 and so x^2 - 6x + 5 = (x-3)^2 - 4 = 0. Now, once we have done this we can rearrange to get (x-3)^2 = 4, so x-3 is plus or minus 2, so x is 1 or 5. This method is really powerful because if you get that l is negative, so you get a square is equal to a negative number, then there are clearly no solutions.

A final (and in my opinion the least elegant way) is to use the quadratic formula for ax^2 + bx + c = 0, which is x = (-b +- sqrt(b^2 - 4ac))/2a. This formula can actually be derived from completing the square.

Answered by Warren L. Maths tutor

2599 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Many students do not understand the rules for when one can 'cancel' in regards to fractions


write log2(5) +log2​​​​​​​(3) in its simplest form


Solve the simultaneous equations: 3x-y=1, x^2+y^2=5


Find the centre and radius of the circle with equation: x^2 + y^2 -4x +8y = 5, and determine whether the point (7,-4) lies on the circle.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy