Question

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

Answer 1

Step-by-step explanation:

Let the required number be x.

$\therefore \: x = 3 \sqrt{x} + 4 \\ \therefore \: x - 4= 3 \sqrt{x} \\ squaring \: both \: sides \\ (x - 4) ^{2} =( 3 \sqrt{x})^{2} \\ \therefore \: {x}^{2} - 8x + 16 = 9x \\ \therefore \: {x}^{2} - 8x + 16 - 9x = 0 \\ \therefore \: {x}^{2} - 17x + 16 = 0 \\ \therefore \: {x}^{2} - 16x - x+ 16 = 0 \\ \therefore \: x(x - 16) - 1(x - 16) = 0 \\ \therefore \: (x - 16)(x - 1) = 0 \\ \therefore \: (x - 16) = 0 \: or \: (x - 1) = 0 \\ \therefore \:x = 16 \: or \: x = 1 \\ \because \: x = 1 \: does \: not \: satisfy \: the \: equation \\ \therefore \: x \: neq \: 1 \\ \huge \red{\boxed{ \therefore \: x = 16}}$