Solve x^2-5*x+6=0

There are 2 methods we can use here.

Method 1 - Factorisation

If we can rewrite the equation so that it looks like (x+a)*(x+b)=0, where 'a' and 'b' are numbers, then we can easily find a solution. If one or both of the terms inside the brackets equals 0, then their product is also 0. This means the solution(s) are the values of x such that x+a=0 or x+b=0.

Multiplying out the brackets of (x+a)(x+b)=0 gives x2+(a+b)x+ab=0. If we match up the terms with the equation we want to solve (called comparing coefficients), we see that we want two numbers 'a' and 'b' so that a+b=-5, and ab=6. The two numbers which satisfy this are -2 and -3, so these are our values for 'a' and 'b'.

Now we can write our equation as (x-2)(x-3)=0 (you can check yourself this can be multiplied out to give the original equation!)

If x-3=0 then x=3, and if x-2=0 then x=2, so these are our two solutions.

Method 2 - Quadratic Formula

This method can be used when you can't easily 'see' the numbers to use to factorise the equation.

The quadratic formula states that for an equation of the form ax2+bx+c=0, where 'a', 'b', and 'c' are numbers, then

x=(-b+-sqrt(b2-4ac)/(2*a)

For our equation a=1, b=-5 and c=6, which can then be substituted into the formula. Again you can check yourself that it simplifies to give the same answers as method 1

HD
Answered by Hannah D. Maths tutor

7784 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

The diagram shows a sector ORST of a circle with centre O. OR = OT = 10.4 cm. Angle ROT = 120°. (a) Calculate the length of the arc RST of the sector (3s.f)


expand and simplify (x+4)(x-2)^2


There is a right angled triangle, you know the length of the hypotenuse (6) and one other side (3), can you calculate the third side of the triangle?


Write the following in their simplest form: a) 8^0 b) 64^(2/3) c) 0.54545454...


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning