How can you solve any quadratic equation?
1 Answer
The most general methods are the quadratic formula and completing the square.
Explanation:
The most general methods which will cope with any quadratic equation in one variable are:

The quadratic formula.

Completing the square.
These methods are both capable of finding the roots regardless of whether they are integers, rational, irrational or even nonReal Complex numbers.
Quadratic formula
The roots of
#x = (b+sqrt(b^24ac))/(2a)#
Completing the square
Given:
#ax^2+bx+c = 0#
Note that:
#a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)#
So our equation can be rewritten:
#a(x+b/(2a))^2 = b^2/(4a)c#
Dividing both sides by
#(x+b/(2a))^2 = b^2/(4a^2)c/a#
Hence:
#x+b/(2a) = +sqrt(b^2/(4a^2)c/a)#
Which can be simplified to:
#x = (b+sqrt(b^24ac))/(2a)#
Hmmm... that looks familiar.
So completing the square and the quadratic formula are kind of the same thing, but in particular cases completing the square can be a little cleaner:
For example, factoring
#x^2+4x21#
#=x^2+4x+425#
#=(x+2)^25^2#
#=((x+2)5)((x+2)+5)#
#=(x3)(x+7)#
Or using the quadratic formula:
#x^2+4x21# is#ax^2+bx+c# with#a=1# ,#b=4# ,#c=21#
hence has zeros:
#x = (4+sqrt(4^2(4*1*(21))))/(2*1)#
#=(4+sqrt(16+84))/2#
#=(4+sqrt(100))/2#
#=(4+10)/2#
#=2+5#
i.e.
Hence factors:
#(x+7)(x3)#