For what values of #b# is #x^2+bx-2# factorable ?

1 Answer
Apr 26, 2017

Answer:

For integer coefficients #b=+-1#.

For rational coefficients #b=(p^2-2)/p# for any non-zero rational number #p#.

For real coefficients any real value of #b# will work.

Explanation:

Given:

#x^2+bx-2#

Note that this is in the form #ax^2+bx+c# with #a=1# and #c=-2#

This quadratic has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = b^2-4(color(blue)(1))(color(blue)(-2)) = b^2+8#

#color(white)()#
Integer coefficients

The only integer values of #b# for which #Delta = b^2+8# is a perfect square are #b=+-1#, resulting in the factorable quadratics:

#x^2+x-2 = (x+2)(x-1)#

#x^2-x-2 = (x-2)(x+1)#

#color(white)()#
Rational coefficients

If we permit rational coefficients and rational values of #b# then there are some more possibilities.

For example, if #p# is any non-zero rational number then:

#(x+p)(x-2/p) = x^2+(p-2/p)x-2#

So #b=p-2/p=(p^2-2)/p# is a possibility for any non-zero rational number #p#.

#color(white)()#
Real coefficients

If we permit irrational coefficients, then we just require #Delta >= 0#, which is already true for any real value of #b#.

Then:

#x^2+bx-2 = (x+b/2)^2-(b^2/4+2)#

#color(white)(x^2+bx-2) = (x+b/2)^2-(sqrt(b^2/4+2))^2#

#color(white)(x^2+bx-2) = ((x+b/2)-sqrt(b^2/4+2))((x+b/2)+sqrt(b^2/4+2))#

#color(white)(x^2+bx-2) = (x+b/2-sqrt(b^2/4+2))(x+b/2+sqrt(b^2/4+2))#