Where does the quadratic formulae come from?
The general form of a quadratic equation is, ax2 +bx + c = 0. If we divide all terms by a we get, x2 +(b/a)x + c/a = 0. then by completing the square we get (x+(b/2a))2 + c/a - b2/4a2 = 0 which rearanged is, (x+(b/2a))2 = b2/4a2 - c/a. We can combine the 2 terms on the left hand side of this equation into, b2-4ac which gives the overall equation,
(x+(b/2a))2 = (b2-4ac) / 4a2. If we then square root both sides we have x+b/2a = sqrt(b2-4ac)/2a. By rearanging again we get x= (sqrt(b2-4ac)-b) / 2a. Which looks like the quadratic formulae, there is a subtlety which is that a square root can be either + or -, i.e. the sqrt(4) is 2 or -2 therefore for our quadratic formulae we have to consider both the postivie and negative terms.
Therefore our overall result reduces to the formulae you are familiar with x= (-b ± sqrt(b2-4ac)) / 2a
Related Maths A Level answers
All answers ▸Let y be a function of x such that y=x^3 + (3/2)x^2-6x and y = f(x) . Find the coordinates of the stationary points .
Find the first 3 terms, in ascending powers of x, of the binomial expansion of (2 – 9x)^4 giving each term in its simplest form.
A curve has equation x^2 + 2xy – 3y^2 + 16 = 0. Find the coordinates of the points on the curve where dy/dx =0
Find the exact value of the gradient of the curve y = e^(2- x)ln(3x- 2). at the point on the curve where x = 2.