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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

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)