Question

Find two positive consecutive

even integers such that the
square of the smaller integer
decreased by five times the
larger integer is 536.

Answer 1

`26` and
`28`.

Explanation:

Let
`x` denote the smaller one of the two even integers (
`x > 0` since both integers are positive.) The larger one of the two consecutive even integer would be
`(x + 2)`.

The square of the smaller integer would be
`x^(2)`.

Five times the larger integer would be
`5\, (x + 2)`.

Subtract five times the larger integer from the square of the smaller integer to get
`(x^(2) - 5\, (x + 2))`.

The value of this expression should be equal to
`536`. In other words:

`x^(2) - 5\, (x + 2) = 536`.

Rewrite and simplify this quadratic equation:

`x^(2) + (-5)\, x + (- 546) = 0`.

• 
  `a = 1`.
• 
  `b = (-5)`.
• 
  `c = (-546)`

Apply the quadratic formula to find possible values of
`x`:

`\begin{aligned}x_(1) &= \frac{-b + \sqrt{b^(2) - 4\, a\, c}}{2\, a} \\ &= \frac{-(-5) + \sqrt{(-5)^(2) - 4 * 1 * (-546)}}{2}\\ &= (5 + √(2209))/(2) \\ &=(5 + 47)/(2) \\ &= 26\end{aligned}`.

`\begin{aligned}x_(2) &= \frac{-b - \sqrt{b^(2) - 4\, a\, c}}{2\, a} \\ &= (5 - √(2209))/(2) \\ &=(5 - 47)/(2) \\ &= -21\end{aligned}`.

Since
`x > 0` (both numbers are supposed to be positive),
`x = 26` would be the only valid solution.

Therefore, the two integers would be
`x = 26` and
`x + 2 = 28`.