How do you use the discriminant to find the number of real solutions of #v^2+4v-3=0#?

1 Answer
Binayaka C.
Dec 25, 2016

Roots are real : #r_1=-2 +sqrt7# and #r_2=-2 -sqrt7#.

Explanation:

#v^2+4v-3=0# here #a=1; b=4; c=-3#. Let the discriminant is #D=b^2-4ac = 4^2-2*1*(-3)=16+12=28# Since #D>=0# two roots are real.
Roots are #r_1 = -b/(2a)+ sqrt(D)/(2a) and r_2 = -b/(2a)- sqrt(D)/(2a) #

#:.r_1= (-4)/(2*1)+ sqrt 28/(2*1) = -2 + (cancel2sqrt7)/cancel2 = -2 +sqrt7# and
#:.r_2= (-4)/(2*1) - sqrt 28/(2*1) = -2 - (cancel2sqrt7)/cancel2 = -2 - sqrt7# [Ans]

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