#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# let a, b and c are unequal real positive numbers such that 2b = a + c, then roots of ax^2 + 2bx + c = 0 are(i) real and equal (ii) Real and distinct (iii) imaginary (iv) nothing definite can be saidPlz explain 10 years ago

Dear student,

2b=a+c

Sum of the roots=-2b/a

Product=c/a

This shows the roots are real and distinct...

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

10 years ago

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

10 years ago

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