IRRational Roots

The no. of irrational solutions of the equation -

\sqrt{x^2 + \sqrt{x^2 + 11}} + \sqrt{x^2 - \sqrt{x^2 + 11}} = 4
are - a) 0 b) 2 c) 4 d) Infinite

12 Answers

62
Lokesh Verma ·

Wow this one was a googley :P

I spent atleast 10 minutes thinking on this one..

The answer is zero because the part \sqrt{x^2-\sqrt{x^2+11}} is never real :D

While the other part is always real for any real value of x..

Thus there is no real root for the equation.

Thus there is no rational root :P

19
Debotosh.. ·

where did u practice to face these googlies so well? MCG or LORDS? .....;)

62
Lokesh Verma ·

@ targetiit.com

[3]

341
Hari Shankar ·

x = \pm \sqrt 5 are solutions. (symmetry and monotonic behaviour for x>0 mean no other solutions exist)

If under the square-root there were x4 then we could have called it googley!

62
Lokesh Verma ·

Oops [3] [1]

Itne din baad I gave a couple of answers today and those too with mistakes [2]

13
Avik ·

But sir, how do v go about finding its solutions ?

V can take x2 + 11 =y2 so tht it becomes -

y2 + y -11 + √y2 - y -11 = 4

Wat to do after tht ?....

13
Avik ·

Waise, it will solved by the usual Shifting n Squaring method [3] .... but a shorter one wud be better...

62
Lokesh Verma ·

I was thinking of rationalizing and getting two simultaneous equations...

but that does not seem to help a lot.

13
Avik ·

Haan, yes bhaiya !....Rationalisation helped too but tht solution is nearly of the same length...

But v get x = ± √5 ...then why is the ans given to be - c) 4 ....?

62
Lokesh Verma ·

but that is not a rational root!

341
Hari Shankar ·

wait, is this function not increasing for x>0? How come four roots then?

13
Avik ·

Confusion not solved yet....!!

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