How do you find the value of the discriminant and state the type of solutions given #-2x^2-x-1=0#?

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Answer 1 · 2017-10-18T19:05:58

no real roots; #Delta<0#

#-2x^2-x-1=0# is already in #ax^2+bx+c# form, so the #a#,#b# and #c# -values can be used.

#a=-2#
#b=-1#
#c=-1#

#Delta=b^2-4ac#
#=(-1)^2 - (4*(-2*-1))#
#=1-8#
#-7#

#-7<0 therefore Delta<0#

hence, #-2x^2-x-1=0# has no real roots.

this can also be seen by graphing #-2x^2-x-1=0#:

this parabola does not meet the #x#-axis on a graph for real numbers, so there are no (real) roots.