Question

How do you factor and solve `x^2-6x-5=0`?

Answer 1

Use the quadratic formula to find

`x^2-6x-5=(x-3-sqrt(14))(x-3+sqrt(14))=0`

Explanation:

The first thing we notice is that `5` is a prime number, so the only integer solutions which multiply to give `5` would be `+-1` and `+-5`, however, the only way to get these to add to be `-6` is if they are both negative, which results in `+5`, therefore, there are no integer solutions to factor.

The solution now is to use the quadratic formula to get the zeros of our equation and substitute them into the factored equation:

`(x-p)(x-q) = x^2-6x-5=0`

Where we find `p` and `q` from

`p,q=(-b+-sqrt(b^2-4ac))/(2a)=(6+-sqrt(36+20))/(2)=3+-sqrt(14)`

so our equation becomes:

`x^2-6x-5=(x-3-sqrt(14))(x-3+sqrt(14))=0`