Answer:
The two consecutive even integers = (4, 6)
Explanation:
We are asked in the question to find two consecutive even integers such that 5 times their sum is 26 more than their product.
Two consecutive even integers is represented as: x, x + 2
Five times their sum is 26 more than their products
5( x + x + 2) = [x × (x + 2)] + 26
5( 2x + 2) = (x² + 2x) + 26
10x + 10 = (x² + 2x ) + 26
x² + 2x + 26 - 10x - 10 = 0
x² - 8x + 16 = 0
We factorise
x² - 4x - 4x + 16 = 0
(x² - 4x) -(4x + 16) = 0
x(x - 4) -4(x - 4) = 0
(x - 4)(x - 4) = 0
(x - 4)²
Hence
x - 4 = 0
x = 4
Since, we know that two consecutive even integers = (x , x + 2)
First even integer = x = 4
Second even integer = x + 2 = 4 + 2
= 6
Therefore, the two consecutive even integers = (4, 6)