Question

How do you solve `x^2 -8x-7=0` by completing the square?

Answer 1

`x= 4+-sqrt(23)`

Explanation:

To complete the square we note that:

`(x-a)^2 -= x^2 - 2ax + a^2`

So the value of `a` is `1/2` the coefficient of `x`.

Thus:

`x^2 -8x -7 -= (x-4)^2 - (4)^2 - 7`
`" " = (x-4)^2 - 16 - 7`
`" " = (x-4)^2 - 23`

And so to solve the equation we have:

`\ \ \ \ \ \ \ x^2 -8x -7 = 0`

`:. (x-4)^2 - 23 = 0`
`:. (x-4)^2 = 23`
`:. x-4 = +-sqrt(23)`
`:. x= 4+-sqrt(23)`

Answer 2

`x = 4 + sqrt23`

Explanation:

`x^2 - 8x - 7` is almost a perfect square.

It would actually be a perfect square if it had been
`x^2 - 8x + 16`, which is a perfect square.

But in order to change the given expression into the perfect square,
you have to add `23` to bring `-7` up to `16`

`x^2 - 8x - 7 = 0`
Solve for `x`

1) Add `23` to both sides to complete the square
`x^2 - 8x - 7 + 23 = 23`

2) Combine like terms
`x^2 - 8x  + 16 = 23`

3) Write the trinomial as a perfect square
`(x-4)^2 = 23`

4) Find the square roots of both sides
`x - 4 = sqrt23`

5) Add `4` to both sides to isolate `x`
`x = 4 + sqrt23` `larr` answer