Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Answer :-

Required natural number is 9.

Explanation :-

Let the natural number be 'x'

Natural number 'x' = 3 more than twice the square root of the natural number 'x'

⇒ x = 2√x + 3

⇒ x - 3 = 2√x

Squaring on both sides

⇒ (x - 3)² = (2√x)²

⇒ (x)² - 2(x)(3) + (3)² = 2²(√x)² [Since (a - b)² = a² - 2ab + b²]

⇒ x² - 6x + 9 = 4(x)

⇒ x² - 6x + 9 = 4x

⇒ x² - 6x + 9 - 4x = 0

⇒ x² - 10x + 9 = 0

Splitting the middle term

⇒ x² - 9x - x + 9 = 0

⇒ x(x - 9) - 1(x - 9) = 0

⇒ (x - 9)(x - 1) = 0

⇒ x - 9 = 0 or x - 1 = 0

⇒ x = 9 or x = 1

If x = 1

⇒ x = 2√x + 3

⇒ 1 = 2(√1) + 3

⇒ 1 = 2(1) + 3

⇒ 1 ≠ 2 + 3

So x ≠ 1

If x = 9

⇒ x = 2√x + 3

⇒ 9 = 2(√9) + 3

⇒ 9 = 2(3) + 3

⇒ 9 = 6 + 3

⇒ 9 = 9

So x = 9

Therefore the required natural number is 9.