Question

the square of the greater of two consecutive even numbers exceeds the square of the smaller number by 36. what are the numbers?

Answer 1

^^Hello!!!^^

Here's The Answer :

Let the smaller number be x.

=>Next consecutive even number after
x .
( x + 2 )

=>According To The Question.

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

${x}^{2}$ + 4x + 4 - ${x}^{2}$ = 36

${x}^{2}$ - ${x}^{2}$ + 4x + 4 = 36

4x + 4 = 36

4x = 36 - 4

4x = 32

x = 32 /4

x = 8

So, the smaller number is 8 .

=>Next consecutive even number after 8 .

=(8 + 2)
=10

Hope It Helps!!!!

Answer 2

$\huge {\boxed {\fcolorbox{pink}{red} {\purple{ur\:answer}}}}$

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

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

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

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

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

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

$2x + 1 = 36$

$2x = 36 - 1 = 35$

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

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