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 .