Question

How do you solve `(x-3)^2 = 5`?

Answer 1

See a solution process below:

Explanation:

First, put the equation in standard quadratic form:

`(x - 3)^2 = 5`

`(x - 3)(x - 3) = 5`

`x^2 - 3x - 3x + 9 = 5`

`x^2 - 6x + 9 = 5`

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

`x^2 - 6x + 4 = 0`

We can now 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)(4)` for `color(green)(c)` gives:

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

`x = (6 +- sqrt(36 - 16))/2`

`x = (6 +- sqrt(20))/2`

`x = (6 - sqrt(4 * 5))/2`; `x = (6 + sqrt(4 * 5))/2`

`x = (6 - sqrt(4)sqrt(5))/2`; `x = (6 + sqrt(4)sqrt(5))/2`

`x = (6 - 2sqrt(5))/2`; `x = (6 + 2sqrt(5))/2`

`x = 6/2 - (2sqrt(5))/2`; `x = 6/2 + (2sqrt(5))/2`

`x = 3 - sqrt(5)`; `x = 3 + sqrt(5)`

Answer 2

`x=3+sqrt5` and `x=3-sqrt5`

Explanation:

Let's take the square root of both sides. This gives us

`x-3=+-sqrt5`

We can add `3` to both sides to get

`x=3+sqrt5` and `x=3-sqrt5`

Hope this helps!