How do you complete the square to solve #-x^2-2x+3= 0#?

2 Answers
Jun 10, 2015

Answer:

#(x+1)^2-4=0#
Solutions: #x=1 or x=-3#

Explanation:

If you have to solve the equation you can multiply all by -1:
#−x^2−2x+3=0 -> x^2+2x-3=0#
It's always more simple to work with positive numbers =)
# x^2+2x-3=0# is similar to a perfect square trinomial that is #x^2+2x+1=(x+1)^2#.
So we can write our equation adding (or subtracting) what is missing to obtain a perfect square:
# x^2+2x-3=(x+1)^2-4=0#
If you want to solve the equation you have to obtain 4 in the square #(x+1)^2=4#, so #x+1=+-2# and the solutions are:
#x=1 or x=-3#

Jun 11, 2015

Answer:

#x=1, -3#

Explanation:

A perfect square trinomial is in the form #(a+b)^2=a^2+2ab+b^2#.

#-x^2-2x+3=0#

Multiply the equation times #-1#.

#x^2+2x-3=0#

The left side of the equation can be made into a perfect square trinomial by completing the square. This will enable the equation to be solved for #x#.

Add #3# to both sides.

#x^2+2x=3#

Divide the coefficient of the #x# term by #2#, then square the result. Add it to both sides of the equation.

#(2/2)^2=1#

#x^2+2x+1=4#

There is now a perfect square trinomial on the left side of the equation.

#(a+b)^2=a^2+2ab+b^2#

#a=x#
#b=1#

#(x+1)^2=4#

Take the square root of both sides.

#x+1=+-sqrt4#

#x+1=+-2#

If #x+1=2#, #x=1#.

If #x+1=-2#, #x=-3#