How can you solve any quadratic equation?

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Answer 1 · 2016-05-23T20:34:39

The most general methods are the quadratic formula and completing the square.

The most general methods which will cope with any quadratic equation in one variable are:

These methods are both capable of finding the roots regardless of whether they are integers, rational, irrational or even non-Real Complex numbers.

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Quadratic formula

The roots of #ax^2+bx+c = 0# are given by the formula:

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

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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 #a# we find:

#(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^2-4ac))/(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+4x-21# by completing the square:

#x^2+4x-21#

#=x^2+4x+4-25#

#=(x+2)^2-5^2#

#=((x+2)-5)((x+2)+5)#

#=(x-3)(x+7)#

Or using the quadratic formula:

#x^2+4x-21# 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. #x=-7# and #x=3#

Hence factors:

#(x+7)(x-3)#