5. A natural number is greater than twice
its square root by 3. Find the number.
Sol. Let the natural number be x,
As per the given condition,
x = 2√x + 3
∴ x – 3 = 2√x
Squaring on both the sides,
(x – 3)2 = (2√x)2
∴ x2 – 6x + 9 = 4 x
∴ x2 – 6x – 4x + 9 = 0
∴ x2 – 10x + 9 = 0
∴ x2 – 9x – x + 9 = 0
∴ x(x – 9) – 1(x – 9)= 0
∴ (x – 1) (x – 9) = 0
∴ x – 1 = 0 or x – 9 = 0
∴ x = 1 or x = 9
Here, both the answers are positive
numbers.
∴ both will be acceptable as answer,
But, if we accept x = 1, then it will contradict
the given condition of A natural number is greater than twice its square root by
3 , Let us check it out,
As per given condition,
x = 2√x
+ 3
If x = 1, then
LHS = x = 1 ...... (i)
RHS = 2√x + 3 = 2√1 + 3 = 2(1) + 3 = 2 + 3
= 5 .......(ii)
From (i) and (ii)
LHS ≠ RHS
∴ we can assume only one value of x
i.e. x = 9, which satisfies the given
condition of A natural number is greater than
twice its square root by 3,
∴ the natural number is 9.
