Question

a^½+a^-½/1-a + 1-a^-½/1-√a. Class 9. Only answer if you know otherwise get lost. Don't spam​

Answer 1

$\Huge{\boxed{\mathbb{\fcolorbox{blue}{purple}{HELLO!!}}}}$

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.

i hope it helps you................

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