Question

FIND THE ANSWER FOR THE
QUESTION B

Answer 1

Answer:

                      Here you go mate.

Step-by-step explanation:

Given :

           N < 100 ,

         √N and ∛N are integers.

a) To prove:

N is a square number.

Let us assume N to be a non square number.

They have given √N is an integer we know only root of a square number will result in an integer when square rooted.

So the statement √N is an integer contradicts our assumption and hence our assumption was wrong.

If our assumption of saying it a non square is wrong then it is an square number.

Hence proved.

b) Root of an integer will not be a real number as square is formed when a number is multiplied to itself. When we consider the signs ( + × + ) , ( - × - ) both results in positive and so there cannot be any perfect square with negative sign. So N is a positive number and is less than 100.

As N is both a perfect square and perfect cube N can be written in form of$x^{6}$ .

Lets try with 0:

                        $0^{6}$ = 0

As 0 < 100 , 0 can be a solution of N.

Lets try with 1:

                      $1^{6}$ = 1

As 1 < 100 , 1 can be a solution of N.

Lets try with 2:

                       $2^{6}$ = 64

64 < 100 , 2 can be a solution of N.

Lets try with 3:

                      $3^{6}$ = 729

As 729 > 100 , 3 cannot be a solution of N.

If $3^{6}$ even exceeds 100 definitely power 6 of other succeeding number will be greater than 100.

And so N can be 0 , 1 , 2.

                                            Answer: 0 , 1 , 2 .

Hope it helps.

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