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?

1 Answer
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