Dear student,

2b=a+c

Sum of the roots=-2b/a

Product=c/a

This shows the roots are real and distinct...

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

for any quadratic eq  ax²+bx+c roots are given by

 X = {-b+_ D)/2a

D =discriminant = (b²-4ac)^1/2

our eq is ax²+2bx+c so

D = {(2b)²-4ac)^1/2 ........1

2b = a+c  (given)

so , (2b)² = (a+c)² ...........2

putting  eq2 in eq1

D = { (a+c)² - 4ac}^1/2  =  { a²+c²-2ac}^1/2 ={(a-c)²}^1/2 =a-c

now roots are

 X = (-2b +_ D}/2a = {-2b + (a-c)}/2a     or        {-2b-(a-c)}/2a                                   

                           = -c/a                      or        -1                                  (on putting 2b =a+c )

so the roots are unequal & real

option (ii) is correct

the equation is ax^2+2bx+c=0

the discriminant is= B^2-4AC

A=a,B=2b,C=c

discriminant= 4b^2-4ac

but 2b=a+c, therefore (2b)^2=(a+c)^2

= (a+c)^2-4ac

=a^2+c^2+2ac-4ac

=a^2+c^2-2ac

=(a-c)^2, which is always greater than zero

therefore roots are real and distinct