Can you use square roots to solve all quadratic equations?

1 Answer
Dec 22, 2014

If the question is about using the square root directly against the equation, the answer is definitely NO.

However, with certain transformation of a given equation into a different but equivalent form it is possible. Here is the idea.

Assume, for example, the same equation as analyzed in the previous answer:
#x^2+x=63#

If we could transform it to something like #y^2=b# then the square root of both sides would deliver a solution.
So, let's transform our equation to this form.
Expression #x^2+x# is not a square of anything, but #x^2+x+1/4# is a square of #x+1/2# because
#(x+1/2)^2=x^2+2*x*1/2+1/4=x^2+x+1/4#

Therefore, it is reasonable to transform the original equation into
#(x+1/2)^2-1/4=63# or
#(x+1/2)^2=253/4#
From the last equation, which is absolutely equivalent to the original one, using the operation of the square root, we derive two linear equations:
#x+1/2=sqrt(253)/2# and #x+1/2=-sqrt(253)/2#

So, two solutions are:
#x=(-1+sqrt(253))/2# and #x=(-1-sqrt(253))/2#

The above method is pretty universal and handy if you don't remember a formula for solutions of a quadratic equation. Let me illustrate this with another example.
#-3x^2+2x+8=0#

Step 1. Divide everything by #-3# to have #x^2# with a multiplier #1#:
#x^2-2/3x-8/3=0#

Step 2. Since a coefficient at #x# is #-2/3#, use #(x-1/3)^2# in a transformed equation:
#(x-1/3)^2-1/9-8/3=0# or
#(x-1/3)^2=25/9#

Step 3. Use square root:
#x-1/3=5/3# and #x-1/3=-5/3#

Step 4. Solutions:
#x=6/3=2# and #x=-4/3#