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 ▸Find the area contained under the curve y =3x^2 - x^3 between 0 and 3
integrate cos^2(x)*sin(x)
Find the equation of the tangent of the curve y = (8x)/(x-8) at the point (0,0)
Given ∫4x^3+4e^2x+k intergrated between the bounds of 3 and 0 equals 2(46+e^6). Find k.