How do you solve `x^2-10x+9=0`?

To factor you must play with factors which multiply to `9`
`(1xx9, " "3xx3, " " 9xx1)`

`(x - 9)(x - 1) = 0`

Now, solve each factor for `0`:

`x - 9 = 0`

`x - 9 + 9 = 0 +9`

`x - 0 = 9`

`x = 9`

and

`x - 1 = 0`

`x - 1 + 1 = 0 + 1`

`x - 0 = 1`

`x = 1`

Given:

`x^2-10x+9 = 0`

`color(white)()`
Sum of coefficients shortcut

Notice that the sum of the coefficients is `0`.

That is:

`1-10+9 = 0`

Hence `x=1` is a solution and `(x-1)` a factor:

`x^2-10x+9 = (x-1)(x-9)`

There are several ways to spot that the other factor must be `(x-9)`. For example, the coefficient of `x` must be `1` so that when multiplied by the `x` in `(x-1)` results in `x^2` and the constant term must be `-9` so that when multiplied by `-1` gives `+9`.

So the other solution is `x=9`

`color(white)()`
Completing the square

Another method, which is a little over the top for this particular problem, involves completing the square, then using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x-5)` and `b=4` as follows:

`0 = x^2-10x+9`

`color(white)(0) = x^2-10x+25-16`

`color(white)(0) = (x-5)^2-4^2`

`color(white)(0) = ((x-5)-4)((x-5)+4)`

`color(white)(0) = (x-9)(x-1)`

Hence solutions `x=9` and `x=1`

`color(white)()`
Quadratic formula

For completeness, I should also mention the quadratic formula.

The equation:

`x^2-10x+9=0`

is in the form

`ax^2+bx+c=0`

with `a=1`, `b=-10` and `c=9`

This has solutions given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (10+-sqrt((-10)^2-4(1)(9)))/(2*1)`

`color(white)(x) = (10+-sqrt(100-36))/2`

`color(white)(x) = (10+-sqrt(64))/2`

`color(white)(x) = (10+-8)/2`

`color(white)(x) = 5+-4`

That is:

`x = 5+4 = 9" "` or `" "x = 5-4 = 1`