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
[tex]x(x+2)=483[/tex]
Apply distributive property left side
[tex]x^2+2x=483[/tex]
[tex]x^2+2x-483=0[/tex]
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+2x-483=0[/tex]
so
[tex]a=1\\b=2\\c=-483[/tex]
substitute in the formula
[tex]x=\frac{-2(+/-)\sqrt{2^{2}-4(1)(-483)}} {2(1)}[/tex]
[tex]x=\frac{-2(+/-)\sqrt{1,936}} {2)}[/tex]
[tex]x=\frac{-2(+/-)44} {2)}[/tex]
[tex]x=\frac{-2(+)44} {2)}=21[/tex]
[tex]x=\frac{-2(-)44} {2)}=-23[/tex]
so
x=21
x+2=23
therefore
The two positive integers that satisfy the requirement are 21 and 23