# How do you find three consecutive positive integers such that the square of the first, increased by the last, is 22?

Oct 19, 2015

Let's cal the integers $x , x + 1 , x + 2$

#### Explanation:

Then ${x}^{2} + \left(x + 2\right) = 22$. Now subtract $22$ on both sides:

${x}^{2} + x - 20 = 0 \to$ which can be factorized into:

$\left(x + 5\right) \left(x - 4\right) = 0 \to$

So either $\left(x + 5\right) = 0 \to x = - 5$ which is not positive,

Or $x - 4 = 0 \to x = 4$ and this one is positive

Answer : $4 , 5 , 6$

Check : ${4}^{2} + 6 = 22$