Answer:
The equation that can be used to solve for m is
[tex]m^{2}-3m-108=0[/tex]
Step-by-step explanation:
Let
m------> the greater integer
we know that
The 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]
The greater positive integer number is 12, and the other number is 9, and if the two positive integers are 3 units apart on a number line. Their product is 108.
Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.
Let's suppose the first integer is x and the other integer would be (x-3)
Their product is 108.
x(x-3) = 108
Using distributive property:
[tex]\rm x^2-3x = 108[/tex]
[tex]\rm x^2-3x -108=0[/tex] (subtracting 108 on both sides)
[tex]\rm x^2-12x+9x -108=0[/tex]
[tex]\rm x(x-12)-9(x-12)=0[/tex]
[tex]\rm (x-9)(x-12)=0[/tex]
x = 9
x = 12
Thus, the greater positive integer number is 12, and the other number is 9, and if the two positive integers are 3 units apart on a number line. Their product is 108.
Learn more about quadratic equations here:
brainly.com/question/2263981
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