Question 607241
Since it is a right triangle, you can employ the Pythagorean theorem to solve for x: {{{c^2 = a^2+b^2}}} and the longest side (the hypotenuse) is x+2
{{{(x+2)^2 = (x+1)^2+x^2}}} Simplify.
{{{(x^2+4x+4) = (x^2+2x+1)+x^2}}} Combine like-terms.
{{{x^2+4x+4 = 2x^2+2x+1}}} Subtract {{{x^2}}} from both sides.
{{{4x+4 = x^2+2x+1}}} Subtract 4x from both sides.
{{{4 = x^2-2x+1}}} Subtracr 4 from both sides.
{{{0 = x^2-2x-3}}} Factor this trinomial.
{{{(x+1)(x-3) = 0}}} Apply the zero product rule.
{{{x+1 = 0}}} or {{{x-3 = 0}}} so...
{{{x = -1}}} or {{{x = 3}}} Discard the negative solution.
{{{x = 3}}}, {{{x+2 = 5}}}, and {{{x+1 = 4}}}