Intermediate Algebra (6th Edition)

$3,$ $4,$ and $5$ units
Let $x, x+1,$ and $x+2$ be the consecutive numbers. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, then \begin{array}{l}\require{cancel} x^2+(x+1)^2=(x+2)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2+(x^2+2x+1)=x^2+4x+4 \\ 2x^2+2x+1=x^2+4x+4 \\ 2x^2-x^2+2x-4x+1-4=0 \\ x^2-2x-3=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-3)(x+1)=0 .\end{array} Equating each factor to zero (Zero Product Property) then the solutions of the equation above are $x=\{-1,3\} .$ Since $x$ is a measurement, then only $x=3$ is the acceptable solution. Therefore, the consecutive numbers that represent the lengths of the sides of a right triangle are $3,$ $4,$ and $5$ units.