How do you find the roots, real and imaginary, of #y= x^2-x-2 # using the quadratic formula?

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Answer 1 · 2016-09-02T03:41:35

The roots are rational (i.e. real) and are #2# and #-1#.

The quadratic formula gives the roots of the general form of quadratic equation #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#.

The roots of a quadratic equation thus critically depend on the discriminant #b^2-4ac#.

In #y=x^2-x-2#, the discriminant is

#(-1)^2-4×1×(-2)#

= #1+8=9#

As it is a complete square one can easily take square root and roots of equation will be rational (i.e. real).

The roots are

#(-(-1)+-sqrt((-1)^2-4×1×(-2)))/2# or

#(1+-sqrt9)/2=(1+-3)/2#

i.e. #2# and #-1#