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

1 Answer
Jun 6, 2016

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