# Three consecutive natural no.are such that ...

Three consecutive natural no.are such that the squares of the middle no. Exceeds the difference of the square of the other two by 60.find the no.

Plz Plz ans.this

Rashmi Bajpayee 2 years, 12 months ago

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

Then, According to question,

{tex}{\left( {x + 1} \right)^2} = {\left( {x + 2} \right)^2} - {x^2} + 60{/tex}

=>          {tex}{x^2} + 2x + 1 = {x^2} + 4x + 4 - {x^2} + 60{/tex}

=>          {tex}{x^2} - 2x - 63 = 0{/tex}

=>          {tex}{x^2} - 9x + 7x - 63 = 0{/tex}

=>          {tex}x(x-9)+7(x-9)=0{/tex}

=>          {tex}(x-9)(x+7)=0{/tex}

=>          {tex}x=9{/tex} and {tex}x=-7{/tex} (neglected, as it is not a natural number)

Therefore, the three consecutive natural numbers are 9, 10 and 11.

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