Find a natural number whose square

Question:

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Solution:

Let n be a required natural number.

Square of a natural number diminished by 84 = n2 – 84 and thrice of 8 more than the natural number = 3 (n + 8)

Now, by given condition,

$n^{2}-84=3(n+8)$

$\Rightarrow \quad n^{2}-84=3 n+24$

$\Rightarrow \quad n^{2}-3 n-108=0$

$\Rightarrow \quad n^{2}-12 n+9 n-108=0 \quad$ [by splitting the middle term]

$\Rightarrow \quad n(n-12)+9(n-12)=0$

$\Rightarrow \quad(n-12)(n+9)=0$

$\Rightarrow 0 \quad n=12$. $[\because n \neq-9$ because $n$ is a natural number $]$

Hence, the required natural number is 12 .

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