Question

A natural number is greater than thrice it's square root by 4. Find the number.

Answer 1

$\bold{Answer:}$

$\bold{1\ \&\ 16}$

$\bold{Step} - \bold{by} - \bold{step\ explanation:}$

$Let the natural number be \ x^2. \\ \\ [Its square root is also mentioned in the question. So it is better to take \ x^2 ] \\ \\ The number is greater than thrice its square root by 4. So the difference between the number and thrice its square root is 4. \\ \\ \therefore \\ \\ x^2 - 3x = 4 \\ \\ x^2 - 3x - 4 = 0 \\ \\ So we get a quadratic equation. You can solve it by many methods applicable to it. \\ \\ I'm solving it by splitting the middle term.$

$x^2 - 3x - 4 = 0 \\ \\ x^2 - 4x + x - 4 = 0 \\ \\ x(x - 4) + 1(x - 4) = 0 \\ \\ (x + 1)(x - 4) = 0 \\ \\ \therefore\ x = -1\ \ \ ; \ \ \ x = 4 \\ \\ \therefore\ x^2 = 1 \ \ \ ; \ \ \ x^2 = 16 \\ \\ \\ If \ \ x = 4,\ \ 3x = 12 \\ \\ 16 is greater than 12 by 4. \\ \\ \\ If \ x = -1, \ \ 3x = -3 \\ \\ 1 is greater than \ -3\ by 4. \\ \\ Also we took \ x^2\ as the natural number. So there's no mention about \ x\ whether it is positive or negative.$

$\therefore\ \bold{1\ \&\ 16}\ are the answers. \\ \\ \\ Hope this may be helpful. \\ \\ Thank you. Have a nice day. \\ \\ \\ \#adithyasajeevan$