Question

Find two positive integers that satisfy the requirement.

Answer 1

Answer:

The two positive integers that satisfy the requirement are 21 and 23

Step-by-step explanation:

Let

x ----> the first consecutive odd integer

x+2 ---> the second consecutive odd integer

we know that

$x(x+2)=483$

Apply distributive property left side

$x^2+2x=483$

$x^2+2x-483=0$

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

$x^2+2x-483=0$

so

$a=1\\b=2\\c=-483$

substitute in the formula

$x=\frac{-2(+/-)\sqrt{2^{2}-4(1)(-483)}} {2(1)}$

$x=\frac{-2(+/-)\sqrt{1,936}} {2)}$

$x=\frac{-2(+/-)44} {2)}$

$x=\frac{-2(+)44} {2)}=21$

$x=\frac{-2(-)44} {2)}=-23$

so

x=21

x+2=23

therefore

The two positive integers that satisfy the requirement are 21 and 23