Question

A natural number is greater than twice its square root by 3. Find the number.

Answer 1

Let the number be x. Then, by the given condition,
$$\begin{gathered} x=2\sqrt{x}+3----\left(1\right) \\ \Rightarrow x-3=2\sqrt{x} \end{gathered}$$
On squaring both sides, we get:
(x – 3)² = 4x
$\Rightarrow$x² – 6x + 9 = 4x
$\Rightarrow$x² – 10x + 9 = 0
On splitting the middle term –10x as –x – 9x, we get:
x² – x – 9x + 9 = 0
$\Rightarrow$x(x – 1) – 9(x – 1) = 0
$\Rightarrow$(x – 1)(x – 9) = 0
$\Rightarrow$x – 1 = 0 or x – 9 = 0
$\Rightarrow$x = 1 or x = 9
On substituting x = 1 in both sides of equation (1), we get:
L.H.S. = x = 1
R.H.S. = $2\sqrt{1}+3=2+3=5$
L.H.S. $\ne$ R.H.S. So , x = 1 does not satisfy equation (1).

On substituting x = 9 in both sides of equation (1), we get:
L.H.S. = x = 9
R.H.S. = $2\sqrt{9}+3=6+3=9$
L.H.S. = R.H.S. So , x = 9 satisfies equation (1).
Thus, the required number is 9.