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

Question

Eva Abramson · Accepted Answer

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

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)#

Nathan Axel · Answer

#x = -5 pm2sqrt(3) #

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

Remove 13 from each side.

#x^2+10x=-13#

Divide 10 by 2 to get 5. Then, square 5 to get 25. Finally, add 25 to each side:

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

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

It is now a perfect square on the left side:

#(x+5)^2=12#

Subtract both sides' square roots to solve now:

#x+5 = pmsqrt(12)#

Take five away from each side:

#x = -5 pmsqrt(12) #

If desired, simplify the radical:

#x = -5 pm2sqrt(3) #

Jace Anderson · Answer

undefined To solve $x^2 + 10x + 13 = 0$ by completing the square:

1. Move the constant term to the other side.
   $x^2 + 10x = -13$

2. Take half of the coefficient of $x$ (which is 10), square it, and add it to both sides of the equation.
   $x^2 + 10x + (10/2)^2 = -13 + (10/2)^2$
   $x^2 + 10x + 25 = -13 + 25$
   $x^2 + 10x + 25 = 12$

3. Rewrite the left side as a perfect square trinomial and simplify the right side.
   $(x + 5)^2 = 12$

4. Take the square root of both sides.
   $x + 5 = ±√12$

5. Solve for $x$.
   $x = -5 ± √12$

6. Simplify the square root if possible.
   $x = -5 ± 2√3$

Therefore, the solutions are $x = -5 + 2√3$ and $x = -5 - 2√3$.