SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024

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

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.