# How do you find the three consecutive integers such that the sum of the squares of the second and the third exceeds the square of the first by 21?

Mar 8, 2016

-8,-7,-6 or 2,3,4

#### Explanation:

Three consecutive numbers

$n - 1 , n , n + 1$

The sum of the squares of the second and the third exceeds the square of the first by 21:

${n}^{2} + {\left(n + 1\right)}^{2} = {\left(n - 1\right)}^{2} + 21$

${n}^{2} + \left({n}^{2} + 2 n + 1\right) = \left({n}^{2} - 2 n + 1\right) + 21$

${n}^{2} + 4 n - 21 = 0$

$n = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot \left(- 21\right)}}{2}$

$n = \frac{- 4 \pm \sqrt{100}}{2}$

$n = \frac{- 4 \pm 10}{2}$

$n = - \frac{14}{2} = - 7 \mathmr{and} n = \frac{6}{2} = 3$

So you have the solutions:

-8,-7,-6 and 2,3,4