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:The equation that can be used to solve for m is [tex]m^{2}-3m-108=0[/tex]  Step-by-step explanation:Letm------> the greater integerwe know thatThe two positive integers are m and (m-3)so[tex]m(m-3)=108\\m^{2}-3m-108=0[/tex]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]m^{2}-3m-108=0[/tex]  so [tex]a=1\\b=-3\\c=-108[/tex] substitute in the formula [tex]m=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(-108)}} {2(1)}[/tex] [tex]m=\frac{3(+/-)\sqrt{9+432}} {2}[/tex] [tex]m=\frac{3(+/-)21} {2}[/tex] [tex]m=\frac{3+21}{2}=12[/tex] [tex]m=\frac{3-21}{2}=-9[/tex] The greater number is [tex]m=12[/tex]The other number is [tex](m-3)=12-3=9[/tex]