question_answer

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

Answer:

Let the three consecutive natural numbers be \[x,x+1\] and \[x+2\].

According to the given condition,

\[\therefore {{(x+1)}^{2}}-[{{(x+2)}^{2}}-{{x}^{2}}]=60\]

\[{{x}^{2}}+1+2x-[(x+2-x)(x+2+x)]=60\]

\[{{x}^{2}}+2x+1-[2(2+2x)]=60\]

\[{{x}^{2}}+2x+1-4-4x=60\]

\[{{x}^{2}}-2x-63=0\]

\[{{x}^{2}}-9x+7x-63=0\]

\[x(x-9)+7(x-9)=0\]

\[(x+7)(x-9)=0\]

\[\therefore x=9\] or \[-7\]

\[\therefore x=9\]                        (neglect \[x=-7\])

\[\therefore \] Number are 9, 10, 11.