How do you solve #x^2 + 10x + 13 = 0# by completing the square?

2 Answers
May 15, 2016

Answer:

#" "x~~ -8.464" and "-1.536#

Explanation:

Given: #y=x^2+10x+13=0#

This process introduces an error that has to be compensated for. To do this I introduce a corrective as yet unknown value represented by #k#. The value of #k# may be determined after all the other changes have taken place

Compare to #y=ax^2+bx+c#

This then written as a first step as:

#y=a(x^2+b/ax)+c# in your case #a=1#

#color(blue)("Step 1")#

#y=(x^2+10x)+13+k_1 larr" At this point "k_1=0#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

Move the square from #x^2# outside the brackets. This begins to introduce errors.

#y=(x+10x)^2+13+k_2 #

#color(brown)(ul("Just for Evan"))#
#color(brown)(k_2" is whatever is needed to turn "y=(x+10x)^2+13#
#color(brown)("back to "y=x^2+10x+13#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Remove the #x# from the #10x#

#y=(x+10)^2+13+k_3#

#color(brown)(ul("Just for Evan"))#
#color(brown)(k_3" is whatever is needed to turn "y=(x+10)^2+13)#
#color(brown)("back to "y=x^2+10x+13#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

Halve the 10

#y=(x+5)^2+13+k_4#

#color(brown)(ul("Just for Evan"))#
#color(brown)(k_4" is whatever is needed to turn "y=(x+5)^2+13)#
#color(brown)("back to "y=x^2+10x+13#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
NOW WE DETERMINE THE VALUE OF #k#

Just for demonstration lets multiply out the brackets and then compare what we have to the original equation.

#y=x^2+10x color(red)(+25) +13+k_4 larr" New equation"#
#y=x^2+10x color(white)("dd.d")+13 color(white)("ddd.")larr" Original equation" #

For the new equation to work we must have #25+k_4=0 =>color(purple)(k_4=-25)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5 : Substitute for "k_4)#

#color(green)(y=(x+5)^2+13color(purple)(+k_4) color(white)("ddd") -> color(white)("ddd") y= (x+5)^2+13color(purple)(-25) )#

#color(green)(color(white)("dddddddddddddddddddd") ->color(white)("dd")y=(x+5)^2-12 #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Now we solve for "x)#

Set #" " y=0= (x+5)^2-12#

Add 12 to both sides

#12=(x+5)^2#

Square root both sides

#+-sqrt(12)=x+5#

Subtract 5 from both sides

#x=-5+-2sqrt(3)#

Tony B

Jan 1, 2018

Answer:

#x = -5 pm2sqrt(3) #

Explanation:

Start with #x^2+10x+13=0#

Subtract 13 from both sides:

#x^2+10x=-13#

Take 10, divide by 2, to get 5. Square 5 to get 25. Add 25 to both sides:

#x^2+10x +25=-13+25#

#x^2+10x +25=12#

The left side is now a perfect square:

#(x+5)^2=12#

Now solve by taking the square root of both sides:

#x+5 = pmsqrt(12)#

Subtract 5 from both sides:

#x = -5 pmsqrt(12) #

Simplify the radical if you'd like:

#x = -5 pm2sqrt(3) #