Respuesta :

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

ACCESS MORE
EDU ACCESS