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

Question

Vikas TU · Answer

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 · Answer

Dear student The value of a lies between -1/2 and ½ . Hope this clears , feel free to ask questions Good Luck Cheers

Arun · Answer

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)

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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

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

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