Question

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

Answer 1

let the two consecutive even nos be 2x and
2x+2
now by question, where greater no be 2x+2
smaller no be 2x
so (2x+2)^2-(2x)^2=36
or (2x+2+2x)(2x+2-2x)=36
or (4x+2)*2=36
or 4x+2=18
or 4x=16
or x= 4
so the nos are 2*4 and (2*4+2) i.e. 8 and 10

Answer 2

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$let \: no. \: be \: x \: and \: x + 1$

$there \: sq. \: are \: {x}^{2} \: and \: (x + 1 {)}^{2}$

$the \: diff. \: is \: 36$

$= > (x + 1 {)}^{2} - {x}^{2} = 36$

${x}^{2} + 2x + 1 - {x}^{2} = 36$

$\cancel{x}^{2} + 2x + 1 - \cancel {x}^{2} = 36$

$2x + 1 = 36$

$2x = 36 - 1 = 35$

$x = \cancel\frac{{35}}{2} = 17.5$

$so \: the \: no. \: are \: 17.5 \: and \: 18.5$