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Home » 1. The sum of the square of the 2 consecutive natural numbers is 481. Find the numbers.
1. The sum of the square of the 2 consecutive natural numbers is 481. Find the numbers.
CBSE | Class 10 | Class 10 | Frank | Maths | Quadratic Equations
1. The sum of the square of the 2 consecutive natural numbers is 481. Find the numbers.

Solution:-

1. The sum of the square of the 2 consecutive natural numbers is 481. Find the numbers.

Solution:-

let us assume the two consecutive number be

    \[P\]

and

    \[P\text{ }+\text{ }1\]

As per the condition given in the question, sum of the square of the

    \[2\]

 consecutive natural numbers is

    \[481\]

,

    \[{{P}^{2}}+{{\left( P+1 \right)}^{2}}~=\text{ }481\]

    \[{{P}^{2}}+{{\left( P+1 \right)}^{2}}~=\text{ }481\]

Then,

    \[{{P}^{2}}+{{\left( P+1 \right)}^{2}}~=481\]

We know that,

    \[{{\left( a\text{ }+\text{ }b \right)}^{2}}~=\text{ }{{a}^{2}}~+\text{ }2ab\text{ }+\text{ }{{b}^{2}}\]

    \[{{P}^{2}}~+\text{ }{{P}^{2}}~+\text{ }2P\text{ }+\text{ }1\text{ }=\text{ }481\]

    \[2{{P}^{2}}~+\text{ }2P\text{ }+\text{ }1\text{ }=\text{ }481\]

By transposing we get,

    \[2{{P}^{2}}+2P+1-481=0\]

    \[2{{P}^{2}}+2P-480\text{ }=\text{ }0\]

Divide by

    \[2\]

for both side of each term we get,

    \[2{{P}^{2}}/2+2P/2480/2=0/2\]

    \[{{P}^{2}}~+\text{ }P240\text{ }=\text{ }0\]

    \[{{P}^{2}}+16P-15P-240\text{ }=\text{ }0\]

Take out common in each terms,

P(P + 16) - 15(P + 15) = 0

(P + 16) (P - 15) = 0

Equate both to zero,

P + 16 = 0

P - 15 = 0

P = - 16

P = 15

So, P = 15 … [because -16 is not a natural number]

Then,

P = 15

P + 1 = 15 + 1 = 16

Therefore, the 2 consecutive numbers are 15 and 16.

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