How do you find the value of the discriminant and state the type of solutions given #5b^2+b-2=0#?

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Answer 1 · 2017-09-02T10:18:25

#D=41# , two real solutions and # b ~~ -0.74 , b ~~ 0.54#.

#5b^2+b-2 =0 # . Comparing with standard quadratic equation

#ax^2+bx+c=0# we get here # a=5 ,b =1 ,c =-2#

Discriminant # D = b^2-4*a*c = 1 -4*5*(-2)=1+40=41 #

If #D# is positive, we will get two real solutions, if it is zero we get

just one solution, and if it is negative we get complex solutions.

Here #D# is positive , so there are two real solution.

# b= (-b +- sqrt(D))/(2a) or b= (-1+- sqrt41)/10 # or

#:. b ~~ -0.74(2 dp) , b ~~ 0.54(2dp)# [Ans]