Question

How do you solve `(x-8)(x-1)=0`?

Answer 1

One way of seeing this is using some logic and math.

If you have a product that equals zero, as in this case, you must agree that at least one of the factors is zero.

Thus, either `x-8` and/or `x-1` equals zero.

Thus, `x_1=8` and `x_2=1`.

Another way is distributing these factors. To distribute, you need to multiply each term of the first by all the terms in the second, as follows:

`(x-8)(x-1)`

`(x*x)+(x*(-1))+((-8)*x)+((-8)(-1))`
`x^2-x-8x+8`
`x^2-9x+8`

And now let's find its roots. I'll use Bhaskara here.

`(9+-sqrt(81-4(1)(8)))/2`
`(9+-7)/2`
`x_1=(9+7)/2=8`
`x_2=(9-7)/2=1`

:)