Hello from MrBillDoesMath!
Answer: 12
Steps:
Let "n" be the smaller even integer. Then next consecutive even integer is n+2. The questions us to find "n" given that n (n+2) = 168. We can solve this using the quadratic equation or using the approach shown below:
n(n+2) = 168
is equivalent to
n^2 + 2n = 168.
Add 1 to both side to get
n^2 + 2n + 1 = 168 + 1= 169 = 13^2.
But n^2 + 2n +1 = (n+1) ^2, so we have
(n+1)^2 = 13^2
giving
n+1 = 13
or n = 13-1 = 12
Let's check: 12 * (12 +2) = 12 ^ 14 =-168
Regards, MrB
even numbers can be represented as 2n where n is an integer
consecutive even integers are 2 apart (2,4,6,8, etc)
so 2 consecutive even integers can be represeted as 2n and 2n+1
their product is 168, so
(2n)(2n+2)=168
expand/distribute
4n²+4n=168
factor out a 4
4(n²+n)=168
divide both sides by 4
n²+n=42
solve by completing the square
take 1/2 of the linear coefient and square it
the linear coeffient is 1 so 1/2 of 1 is 1/2, and (1/2)^2=1/4
add 1/4 to both sides
n²+n+1/4=42+1/4
factor perfect square trinomial
[tex](n+\frac{1}{2})^2=42.25[/tex]
square root both sides
[tex]n+\frac{1}{2}=\pm 6.5[/tex]
subract 1/2 from both sides
[tex]n=-0.5 \pm 6.5[/tex]
n=-0.5+6.5 or -0.5-6.5
n=6 or -7
if n=6, then 2n=12 and 2n+2=12+2=14
if n=-7, then 2n=-14 and 2n+2=-14+2=-12
so the 2 numbers are 12 and 14 or -12 and -14 (the problem didn't specify the sign of the numbers)
