How to do 22nd question without hit and trial method?

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Answer 1 · 2018-05-03T11:53:34

#3. 2#

Two ways of this:

First way (more complicated):
If #97-x=u# and #x=v#, then we also have #97-v=u# and #x=u#.

Essentially, we can have a solution for #root(4)(97-x)+root(4)(x)=5#, and there is another value of #x# which 'swaps' the two values around.

#root(4)(x)+root(4)(u)=5#
#root(4)(u)+root(4)(x)=5#

This is shown better in the second way of showing this.

Second way :

97 is the sum of two quartic numbers (quartic being something raised to the power of 4, like cubic is to the power of three), 81 and 6.

#81=3^4# and #16=2^4#

#97-16=81#, #97-81=16#

#root(4)(97-16)+root(4)(16)=5#
#root(4)(81)+root(4)(16)=5#
#3+2=5#

#root(4)(97-81)+root(4)(81)=5#
#root(4)(16)+root(4)(81)=5#
#2+3=5#

So, we have a solution for 16 and 81.