How was the quadratic formula obtained.
We want a solution to ax^2+bx+c=0. Complete the square to get a[(x+b/2a)^2 -(b^2)/(4a^2)]+c=0. Expalding brackets and rearanging gets a(x+b/2a)^2=(b^2)/4a -c. Divide by a to get (x+b/2a)^2= b^2/4a^2 -c/a=(b^2-4ac)/4a^2. Then root each side tot get (x+b/2a)=+-(root(b^2-4ac))/2a. Then simply move over the b/2a to get x=-b/2a+- (root(b^2-4ac))/2a=(-b+-(root(b^2-4ac)))/2a.
JH
Related Maths A Level answers
All answers ▸How do you integrate ln(x)?
A particle is moving in the with acceleration (2t - 3) ms^-2 and initial velocity 2ms^-1. Find the distance travelled when the velocity has reached 12ms^-1.
Solve the following equation: x^(3) - 6x^(2) + 11x - 6 = 0
Find the turning point of the line y = x^2 + 2x -1