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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
one month ago

Answers : (3)

Vikas TU
8825 Points
							
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
 
one month ago
Vikas TU
8825 Points
							
Dear student 
The value of a lies between -1/2 and ½ . 
Hope this clears , feel free to ask questions 
Good Luck 
Cheers 
one month ago
Arun
22595 Points
							
 
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)
one month ago
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