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

Find the real values of parameter a such that (2a+1)x^2-a(x-1)=2 has one root greater than 1 and other less than 1

Find  the real values of parameter a such that (2a+1)x^2-a(x-1)=2 has one root greater than 1 and other less than 1

Grade:

3 Answers

Vikas TU
14149 Points
2 years ago
Dear student 
(2a+1)x^2 -a(x-1)= 2
(2a+1)x^2 -ax+a-2 =0 
b =-a 
c = a-2 
a = (2a +1)
Apply roots of quadratic equation formula and solve . 
Put values of a, b , c 
x= a(+or_) sqrt(-7a^2 + 12a +8 ) /(4a+2)  > 1 
sqrt(-7a^2 +12a +8 ) > 4a +2 -a 
Solving this 
8-4 > 9a^2 +7a^2 
4> 16a^2 
1/4 > a^2 
-1/2
So For this a the real values of parameter a such that (2a+1)x^2-a(x-1)=2 has one root greater than 1 and other less than 1
 
Vikas TU
14149 Points
2 years ago
Dear student 
The value of a lies between -1/2 and ½ . 
Hope this clears , feel free to ask questions 
Good Luck 
Cheers 
Arun
25763 Points
2 years ago
 
Dear student
 
for this equation we should use f(1) less than 0
 
hence
 
2a + 1 + 0 – 2 less than 0
 
hence a less than ½
 
or we can say a belongs to ( – infiniy, 1/2)

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