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

Answer:

The value of the lesser integer is $28$

Step-by-step 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

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

$x=\frac{-1(+/-)\sqrt{3249}} {2}$

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

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

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

the solution is

$x=28\\x+1=29$

The numbers are $28$ and $29$