Question

The square of the greater of two consecutive odd number exceeds the square of the smaller 36. find the number
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Answer 1

$\boxed{\bf{\mid{\overline{\underline{\bigstar\: According\: to \: the \: Question :}}}}\mid}$

$\: \: \: \:$

$\begin{gathered}:\implies\tt \Big( x + 2 \Big)^2 = 36 + x^2 \\\\\\:\implies\tt x^2 + 4 + 4x = 36 + x^2 \\\\\\:\implies\tt 4x + 4 = 36 \\\\\\:\implies\tt 4x = 36 - 4 \\\\\\:\implies\tt 4x = 32 \\\\\\:\implies\tt x = \dfrac{32}{4} \\\\\\:\implies\tt x = 8 \end{gathered}$

$\dag\:\underline{\textsf{Here we get value of x is \textbf{8}}}.†$

$\: \: \: \:$

$\begin{gathered}:\implies\tt \Big(x + 2 \Big) \\\\\\:\implies\tt 8 + 2 \\\\\\:\implies\huge\tt 10 \end{gathered}$

$\therefore\:\underline{\textsf{Required Numbers are \textbf{8 \& 10}}}.$

Answer 2

$\huge \blue\star~ \pink{ \underline { \textbf{\textsf{Solution :- }}}}$

➞( x + 2 )² = 36 + x²

➞ x² + 4 + 4x = 36 + x²

➞ 4x + 4 = 36

➞ 4x = 36 - 4

➞ 4x = 32

➞ x = 32/4

➞ x = 8

Here we get the value of x = 8

➞ x + 2

➞ 8 + 2

➞ 10

Required Numbers are 8 and 10