Question

the square of the greater of two consecutive even numbers exceeds the square of the smaller by 36. find the numbers

Answer 1

Let the consecutive even numbers be x and x+1
A/Q,

(x+1)²-x²=36
x²+2*x*2+22-x²=36
x²+4x+4-x²=36
4x+4=36
x=36/4
x=8
Therefore,
One one number is 8
Another number is x+1 i.e 8+1=9

Answer 2

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

$\: \: \: \:$

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

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

$\: \: \: \:$

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

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