Question

A natural number is greater than three times its square root by $4$. Find the number.

Answer 1

Let the natural number is $=x$ then

According to given question.

$x+3\times \sqrt{x}=4$

$\Rightarrow x+3\sqrt{x}=4$

$\Rightarrow x-4=3\sqrt{x}$

Squaring both side and we get,

$\Rightarrow x^2+16-8x=9x$

$\Rightarrow x^2-8x-9x+16=0$

$\Rightarrow x^2-17x+16=0$

$\Rightarrow x^2-(16+1)x+16=0$

$\Rightarrow x^2-16x-1x+16=0$

$\Rightarrow x(x-16)-1(x-16)=0$

$\Rightarrow (x-16)(x-1)=0$

$\Rightarrow x=1,16$ (Natural no.)

Hence, this is the answer.