a^½+a^-½/1-a + 1-a^-½/1-√a. Class 9. Only answer if you know otherwise get lost. Don't spam
Answers
You forgot to mention options in the question. anyways .
Recalling the formulae ,
(a+b)2=(a−b)2+4ab
Using the above formulae in the given question,
(a+1/a)2=(a−1/a)2+4a.(1/a)
=(a−1/a)2+4
Now, (a−1/a)2+4−2(a−1/a)=12
Let (a−1/a)=z.
=z2−2z+4−12=0
=z2−2z−8=0
= z2−(4−2)z−8=0
= z2−4z+2z−8=0
= z(z−4)+2(z−4)=0
= (z−4)(z+2)=0
Either z=4orz=−2
Either a−1/a=4ora−1/a=−2
Taking first expression ,
=a−1/a=4
= (a2−1)/a=4
=a2−4a−1=0
Solving the above quadratic we get
Either a=(4+√(42−4.1.(−1))/2
a=2+√5
Or ,a=(4−√(42−4.1(−1))/2
a=2−√5
From second expression,
=(a−1/a)=−2
=( a2−1)/a=−2
= a2−1=−2a
=a2+2a−1=0
Solving the above quadratic we get,
=a=(−2+√(4−4.1.(−1))/2
=a=−1+√2
= a=(−2−√(4−4.1.(−1))/2
a=−1−√2
So the values of a are (2−√5),(2+√5) , (−1+√2), and (−1−√2) respectively.