Question

the difference between Square
of two consecutive numbers is
14.1. Find the Number.​

Answer 1

Given :

• The difference between Square
• of two consecutive numbers =
• 14.1

To find :

• Two consecutive numbers

Solution :

Let the two consecutive number be x, x + 1.

• First number = x
• Second number = x + 1

According to the question,

⠀⠀⠀⇒ (x + 1)² - (x)² = 14.1

Using identity,

• (a + b)² = a² + b² + 2ab

⠀⠀⠀⇒ x² + 1 + 2x - x² = 14.1

⠀⠀⠀⇒ 1 + 2x = 14.1

⠀⠀⠀⇒ 2x = 14.1 - 1

⠀⠀⠀⇒ 2x = 13.1

⠀⠀⠀⇒ x = 13.1/2

⠀⠀⠀⇒ x = 6.55

The value of x = 6.55

Therefore, the two consecutive number :-

• First number = x = 6.55
• Second number = x + 1 = 6.55 + 1 = 7.55

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⠀⠀⠀⠀⠀⠀⠀⠀⠀Verification ::

As given in the question, the difference between Square of the two consecutive numbers = 14.1

So, let's find the difference between the square of the numbers.

⠀⠀⠀⇒ (7.55)² - (6.55)²

Using identity,

• (a - b)(a + b) = a² - b²

⠀⠀⠀⇒ (7.55 - 6.55)(7.55 + 6.55)

⠀⠀⠀⇒ (1)(14.1)

⠀⠀⠀⇒ 14.1

The between the square of the two consecutive numbers = 14.1

Hence, verified.

Answer 2

$\large\underline{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf \: Let - \begin{cases} &\sf{first \: number \: be \: x} \\ &\sf{second \: number \: be \: x + 1} \end{cases}\end{gathered}\end{gathered}$

$\large \underline{\tt \:{ According \: to \: statement }}$

$\rm :\longmapsto\: {(x + 1)}^{2} - {x}^{2} = 14.1$

$\rm :\longmapsto\:(x + 1 + x)( \cancel{x }+ 1 - \cancel{x}) = 14.1$

$\: \: \: \: \: \: \: \: \: \: \: \{\bf \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \}$

$\rm :\longmapsto\:2x + 1 = 14.1$

$\rm :\longmapsto\:2x = 13.1$

$\rm :\longmapsto\:x = \dfrac{13.1}{2}$

$\rm :\implies\: \boxed{ \bf \: x \: = \: 6.55}$

$\begin{gathered}\begin{gathered}\bf \: Hence - \begin{cases} &\sf{first \: number \: be \: x = 6.55} \\ &\sf{second \: number \: be \: x + 1 = 7.55} \end{cases}\end{gathered}\end{gathered}$