Answer:
The value of the lesser integer is [tex]28[/tex]
Step-by-step explanation:
Let
x-------> the first positive integer (lesser integer)
x+1----> the second positive integer
we know that
[tex]x(x+1)=812[/tex]
[tex]x^{2}+x-812=0[/tex]
Solve the quadratic equation
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +x-812=0[/tex]
so
[tex]a=1\\b=1\\c=-812[/tex]
substitute in the formula
[tex]x=\frac{-1(+/-)\sqrt{1^{2}-4(1)(-812)}} {2(1)}[/tex]
[tex]x=\frac{-1(+/-)\sqrt{3249}} {2}[/tex]
[tex]x=\frac{-1(+/-)57} {2}[/tex]
[tex]x=\frac{-1+57} {2}=28[/tex]
[tex]x=\frac{-1-57} {2}=-29[/tex]
the solution is
[tex]x=28\\x+1=29[/tex]
The numbers are [tex]28[/tex] and [tex]29[/tex]
