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