Show how to derive the quadratic formula

You have a general quadratic of the form: ax^2 + bx + c = 0, where a,b,c are constants (although is consistent for functions). Divide by a (assuming a/=0, this would not be a quadratic in that case):x^2 + (b/a)x + c/a = 0Complete the square on the first 2 terms:(x+(b/2a))^2 - (b/2a)^2Add the 3rd term back on:(x+(b/2a))^2 - (b/2a)^2 + c/a = 0Rearrange to have the x term on its own:(x+(b/2a))^2 = (b/2a)^2 - c/aTake the square root:x+(b/2a) = +/-sqrt{(b/2a)^2 - c/a}Subtract b/2a:x = -b/2a +/- sqrt{(b/2a)^2 - c/a}Putting the right hand side over a common denominator:x = [-b +/- sqrt{b^2-4ac}]/2a

TF
Answered by Tom F. Maths tutor

3341 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The curve C is paramterised by the equations: x = 5t + 3 ; y = 2 / t ; t > 0 Find y in terms of x and hence find dy/dx


How would you solve (2x+16)/(x+6)(x+7) in partial fractions?


Solve the simultaneous equations, 2x+y-5=0 and x^2-y^2=3


Time, T, is measured in tenths of a second with respect to distance x, is given by T(x)= 5(36+(x^2))^(1/2)+4(20-x). Find the value of x which minimises the time taken, hence calculate the minimum time.


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