#### 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

# if the roots of the equation 1/x+a   +  1/x+b   =1/c  are equql in magitude but opposite in signs , then their product is   (a)-  ½ (a squqre+b square)  (b)-  -1/2 ( a square+ b square)  (c)-  1/2ab  (d)-  -1/2ab

noogler
489 Points
6 years ago
aren’t the first and second optins are same?
whatever it may be the correct ans is -1/2(a2+b2)

on solving the given equation into the form ax2+bx+c=0

we get x2+x(a+b+2c)+ab-(a+b)c=0
given the two roots of it are equal in magnitude but opp in sign
i.e,  sum of roots =0
from given eq. sum of roots=-b/a=-(a+b+2c)
-(a+b+2c)=0............................a+b/2=c
substitute c=a+b/2 in product of roots = c/a =ab-(a+b)c
on solving it u will get the ans :-)

Rishi Sharma
one year ago
Dear Student,

solving the given equation into the form ax2+bx+c=0
we get x2+x(a+b+2c)+ab-(a+b)c=0
given the two roots of it are equal in magnitude but opp in sign
i.e, sum of roots =0
from given eq. sum of roots=-b/a=-(a+b+2c)
-(a+b+2c)=0............................a+b/2=c
substitute c=a+b/2 in product of roots = c/a =ab-(a+b)c
on solving it u will get the answer

Thanks and Regards