Question

How do you solve `x^2+6x-5=0` by completing the square?

Answer 1

`x = 0.742 or x = -6.742`

Explanation:

For a quadratic `ax^2 +bx+c=0`, there are specific conditions required to be able to write it as the square of a binomial.

`a=1, " "(b/2)^2 =c`

`x^2 +6x-5=0`

In this case, `a=1`, but `(6/2)^2 = 9 !=-5`

Move `-5` to the other side and add `9` to both sides.

`x^2 +6x color(blue)(+9) = 5color(blue)(+9)`

`(x+3)^2 = 14" "larr` find the square root of both sides.

`x+3 = +-sqrt14`

`x = +sqrt14-3" or "x = -sqrt14-3`

`x = 0.742 or x = -6.742`

Answer 2

`x=-3+-sqrt14`

Explanation:

`"to solve using "color(blue)"completing the square"`

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

`"that is " (6/2)^2=9`

`rArr(x^2+6xcolor(red)(+9))-5=0color(red)(+9)`

`rArr(x+3)^2-5=9`

`"add 5 to both sides"`

`(x+3)^2cancel(-5)cancel(+5)=9+5`

`rArr(x+3)^2=14`

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

`sqrt((x+3)^2)=color(red)(+-)sqrt14larr" note plus or minus"`

`rArrx+3=+-sqrt14`

`"subtract 3 from both sides"`

`xcancel(+3)cancel(-3)=-3+-sqrt14`

`rArrx=-3+-sqrt14`