The product of two consecutive even integers is 288. Find the integers.

Question

The product of two consecutive even integers is 288. Find the integers.

Answer 1

using algebra...

n and n+2 are the integers

n*(n+2)=288

n^2+2n=288

n^2+2n-288=0

288=2*2*2*2*2*3*3=(2*2*2*2)*(2*3*3)=16*18 and 18-16=2

factor

(n+18)(n-16)=0

n+18=0

n=-18

n+2=-16

this is one solution

n-16=0

n=16

n+2=18

this is another solution

as you can see, it's just a matter of factoring 288

using arithmetic...

√288=16.97≈17

16 and 18 are the integers (as well as -16 and -18)

 

Answer 2

It'll help to get a useful equation, because no one feels like multiplying every combination of consecutive even integers until it works. Try to find a pattern between consecutive even integers. It might be useful to write a small list.

 

2, 4, 6, 8, 10...

 

For a pattern of consecutive integers, each term is two more than than the previous term. We'll just call A the smaller term. Because consecutive even integers have a difference of 2, the bigger term is (A + 2). Make an equation out of this now. "Product" means the integers are being multiplied.

 

(A)(A + 2) = 288

A² + 2A = 288

A² + 2A - 288 = 0

 

This is a quadratic equation. Solve it by finding two numbers that add up to the b term, which is 2, but multiply to the c term, which is -288. Those numbers are -16 and 18, so use these magic numbers to factor.

 

(A - 16)(A + 18) = 0

 

Solve for A by setting each term equal to zero.

 

A - 16 = 0

A = 16

A + 18 = 0

A = -18

 

The problem didn't say if the integers has to be positive or negative. The positive combination of numbers would be 16 and 18. The negative would be -16 and -18.