Question

Three consecutive even integers are such that the square of the third is 76 more than the square of the second. Find the three integers.

Answer 1

Let n represent the interger l, the three consective intergers are represented by

$n$

$n + 2$

$n + 4$

The second one represent

$(n + 2) {}^{2} + 76 =( n + 4) {}^{2}$

Simplify both sides

$n {}^{2} + 4n + 4 + 76 = {n}^{2} + 8n + 16$

${n}^{2} + 4n + 4 = {n}^{2} + 8n - 60$

$4n + 4 = 8n - 60$

$4n + 64= 8n$

$64= 4n$

$n = 16$

The intergers are 16,18,20