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

1 Answer
Jun 6, 2016

Answer:

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#