How do you find the discriminant for #0.5x^2-2x=-2# and determine the number and type of solutions?

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Answer 1 · 2017-08-01T11:31:37

#D=0 #, so there is one root and the root is #2#

# 0.5x^2 -2x = -2 or 0.5x^2 -2x +2 =0 #

Comparing with standard quadratic equation #ax^2+bx+c=0#

We get here # a= 0.5 , b = -2 , c=2 #

Discriminant #D= b^2-4ac = 4 - 4*0.5*2 = 4-4= 0# We

know if #D >0 # two real roots , if #D = 0 # one real root. and

if #D < 0 # two complex roots. here #D=0 # so there is one root.

Root : #x = (-b +- sqrt D)/(2a) = 2 /(2*0.5)=2/1=2# [Ans]