How do you solve #x^2 - 4x - 9 = 0# by completing the square?

2 Answers
Jun 6, 2017

Answer:

#x=2+-sqrt13#

Explanation:

#"to complete the square"#

add #(1/2"coefficient of x-term")^2 " to both sides"#

#"that is add " (-4/2)^2=4" to both sides"#

#rArrx^2-4xcolor(red)(+4)-9=0color(red)(+4)#

#rArr(x-2)^2-9=4#

#rArr(x-2)^2=13#

#color(blue)"take the square root of both sides"#

#sqrt((x-2)^2)=+-sqrt13larr" note plus or minus"#

#rArrx-2=+-sqrt13#

#rArrx=2+-sqrt13#

#"or " x~~5.6, x~~-1.6" to 1 decimal place"#

Jun 6, 2017

Answer:

#x = +5.606 or x= -1.606#

Explanation:

#x^2 -4x -9=0#

To complete the square means exactly what it says...

"make an expression into a perfect square by adding the part that is missing..."

The steps are given, refer to the details of the working below:

In #ax^2 +bx +c =0#

#1:rarr" "1x^2-4x" "color(red)(-9)=0#

#2:rarr" "x^2 color(blue)(-4)x" " = color(red)(9)#

#3:rarr" "x^2 -4x " " color(blue)(+4) = 9 color(blue)( +4)#

#4:rarr" "(x-2)^2" " = 13#

#5:rarr" "x-2" " = +-sqrt13#

#6:rarr" "xcolor(white)(wwwwwww) = +sqrt13 +2 = +5.606#
#6:rarr" "xcolor(white)(wwwwwww) = -sqrt13 +2 = -1.606#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

  • Step 1: Make #a=1" " (a# is already equal to #1# )

  • Step 2: Move the constant to the other side.

  • Step 3: complete the square by adding #color(blue)((b/2)^2)# to both sides.
    In this case #color(blue)(b=-4)" "# so, #color(blue)(((-4)/2)^2 = (-2)^2 =+4)#

  • Step 4: Write the LHS as the square of a binomial

  • Step 5: square root both sides #rarr# remember #+-sqrt#
  • Step 6: solve for #x# to get 2 values