Question

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

Answer 1

`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"`

Answer 2

`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