Question

Derive the quadratic formula from the standard form (ax2 + bx + c = 0) of a quadratic equation by following the steps below.

a.Divide all terms in the equation by a.
b.Subtract the constant (the term without an x) from both sides.
c.Add a constant (in terms of a and b) that will complete the square.
d.Take the square root of both sides of the equation.
e.Solve for x.

Answer 1

Everyone should do this derivation, because otherwise the "quadratic formula" is some sort of "magic" LOL...

ax^2+bx+c=0

x^2+bx/a+c/a=0

x^2+bx/a=-c/a

x^2+bx/a+b^2x/(4a^2)=b^2/(4a^2)-c/a

(x+b/(2a))^2=(b^2-4ac)/(4a^2)

x+b/(2a)=±√(b^2-4ac)/(2a)

x=-b/(2a)±√(b^2-4ac)/(2a)

x=(-b±√b^2-4ac)/(2a)

Answer 2

See below.

Explanation:

We are going to take the quadratic formula ax²+bx+c=0

a.Divide all terms in the equation by a.

`(ax^(2) )/(a) +(b)/(a)x+(c)/(a)=0\\\\x^(2) +(b)/(a)x+(c)/(a)  =0`

b.Subtract the constant (the term without an x) from both sides.

`x^(2) +(b)/(a)x+(c)/(a) -(c)/(a)  =(-c)/(a) \\\\x^(2) +(b)/(a)x=-(c)/(a)`

c.Add a constant (in terms of a and b) that will complete the square.

`x^(2) +(b)/(a)x=-(c)/(a)\\\\[tex]x^(2) +(b)/(a)x+(b^(2) )/(4a^(2) )  =-(c)/(a) +(b^(2) )/(4a^(2))\\\\(x+(b)/(2a)) ^(2)  =(-4ac+b^(2) )/(4a^(2) )`

d.Take the square root of both sides of the equation.

`x+(b)/(2a)  =[tex]\\x+(b)/(2a) =\frac{\sqrt{-4ac+b^(2) } }{2a}\\x=\frac{\sqrt{-4ac+b^(2) } }{2a}-(b)/(2a)`

e.Solve for x.

`x=-b+\frac{\sqrt{b^(2)-4ac } }{2a}`