Thank you for registering.

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

Please check your email for login details.
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 a and b are positive numbers , prove that the equation 1/x + (1/x+a) +(1/x+b) =0 has 2 real roots ,one b/w a/3 and 2a/3 and other b/w – 2b/3 and – b/3

If a and b are positive numbers , prove that the equation 1/x + (1/x+a) +(1/x+b) =0 has 2 real roots ,one b/w a/3 and 2a/3 and other b/w – 2b/3 and – b/3

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find the solution of your question below,
\frac{1}{x} + \frac{1}{x + a} + \frac{1}{x + b} = 0
\frac{(x+a)(x+b)+x(x+b)+x(x+a)}{x.(x+a)(x+b)} = 0
3x^{2}+2bx+2ax+ab = 0
f(x) = 3x^{2}+2x(a+b)+ab = 0
\Delta = b^{2}-4ac
\Delta = 4(a+b)^{2}-4.3.ab
\Delta = 4(a^{2}+b^{2}+2ab)-12ab
\Delta = 2a^{2}+2b^{2}+2a^{2}+2b^{2}-4ab
\Delta = 2a^{2}+2b^{2}+2(a-b)^{2}
\Delta \geq 0
-> So equation f(x) has two real roots.
f(x) = 3x^{2}+2x(a+b)+ab
You can easily found the value from here.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free