Question 162007
Let the three consecutive integers be: x, (x+1), and (x+2)
{{{(x+1)^2 + (x+2)^2 = x^2+21}}} "...the sum of the squares of the second and third exceed the square of the first by 21." Simplify this and solve for x.
{{{(x^2+2x+1)+(x^2+4x+4) = x^2+21}}} Combine like-terms.
{{{2x^2+6x+5 = x^2+21}}} Subtract {{{x^2}}} from both sides.
{{{x^2+6x+5 = 21}}} Subtract 21 from both sides.
{{{x^2+6x-16 = 0}}} Solve this quadratic equation by factoring.
{{{(x-2)(x+8) = 0}}} Apply the zero product rule.
{{{x-2 = 0}}} or {{{x+8 = 0}}} so...
{{{x = 2}}} or {{{x = -8}}}
There are two answers to this problem:
x = 2
(x+1) = 3
(x+2) = 4 and...
x = -8
x+1 = -7
x+2 = -6