Question

2. Find three consecutive even integers such that the sum of the product of the first two and

the product of the last two is 648

Answer 1

16,18 and 20

Explanation:

Let n= The third number

(n-2)= second number

(n-4)= first number

Therefore,

• The sum of the product of the first two and that of the product of the last two will be in the form;

n(n-2)+(n-2)(n-4)=648

n²-2n+n²-2n-4n+8=648

2n²-8n+8=648

2n²-8n-648+8=0

2n²-8n-640=0

• Dividing through by 2

n²-4n-320=0

n²-20n+16n-320=0

(n-20)(n+16)=0

n=20 , n=-16

NOTE: n can't be equal to a negative number

hence n=20

third number

n=20

second number

(20-2)

18

first number

(n-4)

(20-4)

=16