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

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

$\: \: \: \:$

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

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

$\: \: \: \:$

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

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