How to use the discriminant to find out what type of solutions the equation has for #2x^2 + x - 1 = 0#?

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Answer 1 · 2015-05-21T21:59:12

#2x^2+x-1# is of the form #ax^2+bx+c# with #a=2#, #b=1# and #c=-1#.

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = 1^2-(4xx2xx-1) = 1+8 = 9 = 3^2#

Since this is positive, the equation #2x^2+x-1 = 0# has two distinct real solutions.

Additionally, since it is a perfect square (and the cofficients of the quadratic are rational), those roots are rational.

In fact, they are given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (-1+-3)/4#

That is, the solutions are #x = -1# and #x = 1/2#.