Question

How do you solve `s^ { 2} - 6s = - 7`?

Answer 1

See a solution process below:

Explanation:

First, add `color(red)(7)` to each side of the equation to write the equation in standard quadratic form:

`s^2 - 6s + color(red)(7) = -7 + color(red)(7)`

`s^2 - 6s + 7 = 0`

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(1)` for `color(red)(a)`

`color(blue)(-6)` for `color(blue)(b)`

`color(green)(7)` for `color(green)(c)` gives:

`s = (-color(blue)((-6)) +- sqrt(color(blue)((-6))^2 - (4 * color(red)(1) * color(green)(7))))/(2 * color(red)(1))`

`s = (color(blue)(6) +- sqrt(color(blue)(36) - 28))/2`

`s = (color(blue)(6) +- sqrt(8))/2`

`s = (color(blue)(6) - sqrt(8))/2` and `s = (color(blue)(6) + sqrt(8))/2`

`s = (color(blue)(6) - sqrt(4 * 2))/2` and `s = (color(blue)(6) + sqrt(4 * 2))/2`

`s = (color(blue)(6) - sqrt(4)sqrt(2))/2` and `s = (color(blue)(6) + sqrt(4)sqrt(2))/2`

`s = (color(blue)(6) - 2sqrt(2))/2` and `s = (color(blue)(6) + 2sqrt(2))/2`

`s = color(blue)(6)/2 - (2sqrt(2))/2` and `s = color(blue)(6)/2 + (2sqrt(2))/2`

`s = 3 - sqrt(2)` and `s = 3 + sqrt(2)`

Answer 2

`s = +sqrt2+3" or " s = -sqrt2+3`

Explanation:

Set the quadratic equation equal to `0`.

`s^2 -6s +7 =0`

This does not factorise because there are no factors of `7` which add to give `6`.

Notice that `a=1 and b` is even.
My choice to solve it would be by completing the square:

`s^2 -6s" "= -7" "larr` move the constant
`s^2 -6s color(blue)(+9)= -7 color(blue)(+9)" "larr` add `color(blue)((b/2))^2` to both sides
`(s-3)^2 = 2" "larr` write as the square of a binomial

`s-3 = +-sqrt2`

`s = +sqrt2+3" or " s = -sqrt2+3`