Question

Find a natural no. whose square diminshed by 84 is equal to thrice of 8 more than the given no.

Answer 1

Hi...☺

Here is your answer...✌

Let the natural number be n

Now
According to the question,

${n}^{2} - 84 = 3 \times (n+ 8) \\ \\ {n}^{2} - 84 = 3n + 24 \\ \\ {n }^{2} - 3n - 84 - 24 = 0 \\ \\ {n}^{2} - 3n - 108 = 0 \\ \\ {n}^{2} - 12n + 9n - 108 = 0 \\ \\ n(n - 12) + 9(n - 12) = 0 \\ \\ (n + 9)(n - 12) = 0 \\ \\ = > n = - 9 \: \: or \: \: n = 12 \\ \\ = > n = 12$

[ By rejecting n = -9
As n is natural number ]

HENCE,

The required natural number is 12

Answer 2

let the no si x

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

$x {}^{2} - 84 = 3x + 24$

$x {}^{2} - 3x - 180 = 0$

$x {}^{2} - 12x + 9x - 108 = 0$

$x(x - 12) + 9(x - 12) = 0$

$(x + 9)(x - 12) = 0$

$x = - 9 \: \: a nd \: \: 12$


Hence 12 is the number


Thanks