Question

Two positive integers are 3 units apart on a number line. Their product is 108. Which equation can be used to solve for m, the greater integer?

Answer 1

You can represent the two integers as m and m-3 as the smaller integer is 3 units smaller than the larger integer. Since they multiply to 108, you can use the equation (m)*(m-3)=108 to solve for m.

Answer 2

Answer:

The equation that can be used to solve for m is

$m^{2}-3m-108=0$  

Step-by-step explanation:

Let

m------> the greater integer

we know that

The two positive integers are m and (m-3)

so

$m(m-3)=108\\m^{2}-3m-108=0$

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

$m^{2}-3m-108=0$  

so

$a=1\\b=-3\\c=-108$

substitute in the formula

$m=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(-108)}} {2(1)}$

$m=\frac{3(+/-)\sqrt{9+432}} {2}$

$m=\frac{3(+/-)21} {2}$

$m=\frac{3+21}{2}=12$

$m=\frac{3-21}{2}=-9$

The greater number is $m=12$

The other number is $(m-3)=12-3=9$