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Sladkaya [172]
3 years ago
13

The sum of two squares of two consecutive natural numbers is 41.Find the numbers​

Mathematics
2 answers:
Arada [10]3 years ago
7 0
<h3><u>AnswEr</u><u> </u><u>:</u></h3>

The 2 unknown numbers are 4 and 5 .

<h3><u>Expla</u><u>nation</u><u> </u><u>:</u></h3>

Let the numbers be x and x + 1.

From the given information,

\dashrightarrow\sf \ \ \ x^2 + (x + 1)^2 = 41 \\  \\  \\

\dashrightarrow\sf \ \ \ 2 {x}^{2}  + 2x + 1 = 41 \\  \\  \\

\dashrightarrow\sf \ \ \ 2 {x}^{2}  + 2x + 1 - 41=0 \\  \\  \\

\dashrightarrow\sf \ \ \  {x}^{2}  + x - 20 = 0 \\  \\  \\

\dashrightarrow\sf \ \ \ (x + 5)(x - 4)=0 \\  \\  \\

\dashrightarrow\sf \ \ \ { \underline{ \boxed{ \bf{ \red{x =  - 5,4}}}}} \ \bigstar \\  \\

But, as we know -5 is not a natural number,

Hence, x = 4 .

<u>So the 2 unknown numbers are</u><u> </u><u>:</u>

  • x = 4 [1st number]

  • x + 1 = 4 + 1 = 5 [2nd number]

<h2><u>━━━━━━━━━━━━━━━</u></h2>
Luda [366]3 years ago
5 0

★ Solution :

Let the number be x and x + 1.

\rule{130}1

\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccording} \: to \: \mathscr {Q} \bf{uestion} ....}}}

:\implies\sf x^2 + (x + 1)^2 = 41 \\\\\\:\implies\sf 2x^2 + 2x + 1 - 41 = 0\\\\\\:\implies\sf 2x^2 + 2x + 40 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + 5x - 4x - 20 = 0\\\\\\:\implies\sf x(x + 5) - 4(x + 5) = 0\\\\\\:\implies\sf (x + 5)(x - 4) = 0\\\\\\:\implies\underline{\boxed{\sf x = -5, 4}}

⠀

We know that -5 is not a natural number. Hence, x = 4.

\therefore\:\underline{\textsf{The two consecutive natural number is  \textbf{4 and 5}}}.

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