To solve for which term of the sequence given by [tex]\( I_n = n^2 - n \)[/tex] equals 72, we need to find the value of [tex]\( n \)[/tex] that satisfies the equation [tex]\( n^2 - n = 72 \)[/tex].
1. Set Up the Equation:
We start by writing down the equation representing the sequence:
[tex]\[
n^2 - n = 72
\][/tex]
2. Form a Quadratic Equation:
To solve for [tex]\( n \)[/tex], we rearrange the equation into standard quadratic form:
[tex]\[
n^2 - n - 72 = 0
\][/tex]
3. Solve the Quadratic Equation:
We can solve this quadratic equation, [tex]\( n^2 - n - 72 = 0 \)[/tex], by factoring, completing the square, or using the quadratic formula, [tex]\( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -72 \)[/tex].
4. Quadratic Formula Application:
Plugging in the values into the quadratic formula, we get:
[tex]\[
n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1}
\][/tex]
Simplifying inside the square root:
[tex]\[
n = \frac{1 \pm \sqrt{1 + 288}}{2}
\][/tex]
[tex]\[
n = \frac{1 \pm \sqrt{289}}{2}
\][/tex]
Since [tex]\( \sqrt{289} = 17 \)[/tex]:
[tex]\[
n = \frac{1 \pm 17}{2}
\][/tex]
5. Calculating the Roots:
This gives us two potential solutions for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{1 + 17}{2} = \frac{18}{2} = 9
\][/tex]
[tex]\[
n = \frac{1 - 17}{2} = \frac{-16}{2} = -8
\][/tex]
6. Verify the Solutions:
The quadratic equation yields two solutions: [tex]\( n = 9 \)[/tex] and [tex]\( n = -8 \)[/tex].
Therefore, the terms of the sequence [tex]\( I_n = n^2 - n \)[/tex] that produce the number 72 are for [tex]\( n = 9 \)[/tex] and [tex]\( n = -8 \)[/tex].
