How do you factor the trinomial #x^2+8xy-36y^2=0#?

1 Answer
Apr 20, 2016

Factors of #x^2+8xy-36y^2=0# are #(x+(4-2sqrt13)y)(x+(4+2sqrt13)y)#

Explanation:

Note that the trinomial #x^2+8xy-36y^2=0# is homogeneous function of degree two and hence we can consider it as a quadratic equation.

The discriminant, here is #8^2-4xx1xx(-36)=64+144=208#, and it is not a complete square and hence we will not have rational factors for this equation.

Using quadratic formula #x/y=(-b+-sqrt(b^2-4ac))/(2a)#,

#x/y=(-8+-sqrt208)/2=-8/2+-sqrt(16xx13)/2=-4+-2sqrt13#

Hence factors of #x^2+8xy-36y^2=0# are #(x+(4-2sqrt13)y)(x+(4+2sqrt13)y)#