How do you solve #x^2 + 6x + 10 = 0# by completing the square?

2 Answers
Apr 3, 2016

Answer:

#(x+3)^2 + 1 -> x = +-i - 3#

Explanation:

Completing the square is a method of getting as close as you can with one set of brackets and adding or subtracting the rest.

It ends up in a form like

#x^2 + ax + b = (x + c)^2 + d#

#c# is always half of #a#, so that when you expand out the brackets you have the right coefficients for #x^2# and #x#.

#a = 6 -> c = 3#
#(x + 3)^2 = x^2 + 6x + 9#

As you can see this is pretty close to the original quadratic. All you need to do is add #1# and it equates perfectly.

#x^2 + 6x + 10 = (x+3)^2 + 1#

To then solve, rearrange like so

#(x + 3)^2 + 1 = 0#
#(x + 3)^2 = -1#
#x + 3 = sqrt(-1) = +-i#
#x = +-i - 3#

Apr 3, 2016

Answer:

There is no solution for the given equation for any #x# in the set of 'Real Numbers'.

#color(blue)(x=-3+-i)#

Explanation:

Consider the standard form of #y=ax^2+bx+c#

The by completing the square we have:

#y=a(x+b/(2a))^2 +c - [(b/(2))^2]#

In your case #a=1# so we have:

#y=(x+b/2)^2+c-[(b/2)^2]#

#=>y=(x+3)^2+10 -(3^2)#

#=> y= (x+3)^2+1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#x_("vertex") = (-1)xx(+3) = -3#
#y_("vertex") =+1#

The coefficient of #x^2# = +1 so the graph is of shape type #uu#

Thus the vertex is a minimum and above the x-axis

Thus there is no solution for #x# at #y=0# where #x in RR#

However their will be a solution for #x in CC# (Complex numbers)

Given that #0=(x+3)^2+1#

#=> sqrt((x+3)^2)=sqrt(-1)#

#=>+-(x+3)= i#

Updated: The #+-# is for the #x" and " 3#. Not for the #i#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Suppose we had "-(x+3)= i#

This becomes: #-x-3=i#

#=> -x=+3+i#
Multiply by (-1)

#color(blue)(=>+x=-3-i)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Suppose we had "+(x+3)= i)#

This becomes: #x=-3+i#

#color(blue)(=>+x=-3+i)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Solution for #x# at #y=0# is

#color(magenta)(x=-3+-i)#

Tony B