Question

The product of two consecutive positive integers is 812. What is the value of the lesser integer?

Answer 1

The value of the lesser integer is 28.

Answer 2

The value of the lesser integer is
`28`

Explanation:

Let

x-------> the first positive integer (lesser integer)

x+1----> the second positive integer

we know that

`x(x+1)=812`

`x^(2)+x-812=0`

Solve the quadratic equation

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2) +x-812=0`

so

`a=1\\b=1\\c=-812`

substitute in the formula

`x=\frac{-1(+/-)\sqrt{1^(2)-4(1)(-812)}} {2(1)}`

`x=\frac{-1(+/-)√(3249)} {2}`

`x=\frac{-1(+/-)57} {2}`

`x=\frac{-1+57} {2}=28`

`x=\frac{-1-57} {2}=-29`

the solution is

`x=28\\x+1=29`

The numbers are
`28` and
`29`