Question #4fa4f

1 Answer
May 14, 2016

Answer:

It is a method to Solve Quadratic Equations .
Please remember that in actual practice unless you're told that you use completing the square method, you will probably not use this method.

Explanation:

Suppose you need to solve the following Quadratic equation by completing the square method (I have selected a problem where coefficient of #x^2# term is #1#):
#x^2+6x+1=0# ......(1)

First step is to keep all terms containing #x# on the LHS and take constant terms (not containing #x#) to RHS of the equation. The equation becomes
#x^2+6x=-1# .....(2)

Now take half of the coefficient of the #x#-term [middle term in equation (1)],
square it, and add that value to both sides of the new equation.

In the example we have coefficient of the #x#-term #=6#.

Half of #6=3#. Square of #3=9#. Therefore adding #9# to both sides of equation (2) we obtain

#x^2+6x+9=-1+9#

#=> x^2+6x+9=8#

On inspection we see that we have made LHS a perfect square which can be written as #(x+3)^2#. Doing so the equation becomes

#(x+3)^2=8#

Take square root of both sides and remember to use both #+ and -# signs on one side. We did it for RHS.

#sqrt((x+3)^2)=sqrt8#
#(x+3)=+-sqrt8#

Solve for #x#
#x=-3+-sqrt8#
or #x=-3+-2sqrt2#

The two roots of the given equation are
#x=-3+2sqrt2# and #x=-3-2sqrt2#