Question

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Three consecutive natural numbers ae such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60

Answer 1

Question: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60.

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ANSWER:

There are three consecutive integers (same as, natural numbers).

let the smallest of the numbers be $x$.

Values of $x$ :

1st no. is $x$
2nd no. is $x+1$.
3rd no. is $x+1+1$  $=x+2$.

Condition: The square of the 2nd no. (middle number) is 60 more than the difference of the squares of other two numbers.

Solution:

$(x+1)^{2}=(x+2)^{2}-x^{2}$

=>  $x^{2}+1= x^{2}+4-x^{2}$

=>  $x^{2}-x^{2}+x^{2}=4-1$

=>  $(x^{2}-x^{2})+x^{2}=4-1$

=>  $0+x^{2}=3$

=>  $x^{2}=3$

=>  $x= \sqrt{3}$


By putting the values of $x$,

1st no. is $x= \sqrt{3}$.
2nd no. is $x+1= \sqrt{3}+1$.
3rd no. is $x+2= \sqrt{3}+2$.

ANSWER: The values of the three consecutive numbers:

1st no. is $x= \sqrt{3}$.
2nd no. is $x+1= \sqrt{3}+1$.
3rd no. is $x+2= \sqrt{3}+2$.

Answer 2

Answer:

CHECK THE ATTACHMENT FOR SOLUTION.

HOPE IT HELPS.