find three consecutive even integers such that the product of the first and second is 4 less than the square of the third.
Three consecutive even integers are:
2 n , 2 n + 2 , 2 n + 4
Product of the first and second is 4 less than the square of the third means:
2 n ∙ ( 2 n + 2 ) = ( 2 n + 4 )² - 4
4 n² + 4 n = 4 n² + 2 ∙ 2 n ∙ 4 + 4 ² - 4
4 n² + 4 n = 4 n² + 16 n + 16 - 4
4 n² + 4 n = 4 n² + 16 n + 12
Subtract 4 n² to both sides
4 n = 16 n + 12
Subtract 16 n to both sides
- 12 n = 12
Divide both sides by - 12
n = - 1
Your numbers are:
2 n , 2 n + 2 , 2 n + 4
- 2 , 0 , 2
Proof:
Product of the first and second is 4 less than the square of the third.
( - 2 ) ∙ 0 = 0
2² = 4
( - 2 ) ∙ 0 = 2² - 4 = 4 - 4
n is the 1st integer
n(n + 2) = [(n + 4)^2] - 4 ... n^2 + 2n + 4 = n^2 + 8n + 16
To find three consecutive even integers satisfying the given condition, let's assume the first even integer as x.
The next consecutive even integer will be x + 2, as the difference between consecutive even integers is always 2.
And the third consecutive even integer will be x + 4.
Now, we can form the equation according to the given condition:
Product of the first and second = (x) * (x + 2)
Square of the third - 4 = (x + 4)^2 - 4
According to the given condition, we can write the equation as:
(x) * (x + 2) = (x + 4)^2 - 4
Expanding the equation:
x^2 + 2x = (x^2 + 8x + 16) - 4
Simplifying the equation:
x^2 + 2x = x^2 + 8x + 12
Rearranging the equation:
x^2 + 2x - x^2 - 8x = 12
Combining like terms:
-6x = 12
Dividing by -6 on both sides:
x = 12 / -6
Simplifying:
x = -2
So, the first consecutive even integer is -2.
The second consecutive even integer is -2 + 2 = 0.
The third consecutive even integer is -2 + 4 = 2.
Therefore, the three consecutive even integers satisfying the given condition are -2, 0, and 2.
If the smallest is x, then you want
x(x+2) = (x+4)^2 - 4
Now just solve for x