Integer Solutions to Inequality

Task

What is the sum of all integer solutions to $1\lt (x-2)^2\lt 25$?

IM Commentary

This task provides a reasonably accessible instance of illustrating cluster (A-REI.B: Solve equations and inequalities in one variable) without being limited to only linear equalities (which are specifically addressed in A-REI.3).

The inequality in this task can be solved in a multitude of ways, including through the use of a table, by manipulating the inequalities, or by making a substitution of variables. The first is the most straightforward and leads to a solution relatively quickly, especially if a little reasoning about magnitude is done beforehand. Students might also use graphing technology: if the function $f(x) = (x-2)^2$ is plotted then the graph could be used to identify the integer values satisfying the inequality. The downside to this method is that some experimentation will need to be done to decide on the appropriate domain (the range is effectively given as we are interested in integer values of $x$ where $1 \lt f(x) \lt 25$). Because of the wide variety of approaches possible, the task might be productively used in a context where students come up with their own solution, and then share strategies in groups or as a whole class.

This task was adapted from problem #8 on the 2012 American Mathematics Competition (AMC) 10B Test. On that test, taken by 35,086 students, the multiple choice answers for the problem had the following distribution:

Choice	Answer	Percentage of Answers
(A)	10	4.72
(B)*	12	38.01
(C)	15	38.40
(D)	19	3.55
(E)	25	2.66
Omit	--	12.65

Of the 35,086 students: 17,169, or 49%, were in 10th grade; 9,928 or 28%, were in 9th grade; and the remainder were below than 9th grade.

Solutions

Solution: 1 Solution Table

We aim to make a table of solutions for the inequality. We start with $x = 0$ and then put in positive values of $x$:

$x$	$(x-2)^2$
0	4
1	1
2	0
3	1
4	4
5	9
6	16
7	25
8	36
9	49

The only values that work for our inequality are $x$ = 0, 4, 5, and 6. As $x$ grows, $(x-2)^2$ will continue to grow so there are no more positive solutions to this inequality. For the negative values, we find

$x$	$(x-2)^2$
-1	9
-2	16
-3	25
-4	36
-5	49

We stopped at -5 as the values of $(x-2)^2$ are repeating the values when we plugged in positive $x$'s and so there will be no more negative solutions. Our solutions are $$ x = -2, -1, 0, 4, 5, 6. $$ Adding this up we get 12.

Solution: 2 Inequalities

We begin by taking the square root of the inequality in question: $$ \sqrt{1} \lt \sqrt{(x-2)^2} \lt \sqrt{25}. $$ The square root function preserves the order of numbers larger than 1: that is, if $y$ and $z$ are larger than 1 then $y \gt z$ exactly when $\sqrt{y} \gt \sqrt{z}$. So this means that $1 \lt (x-2)^2 \lt 25$ exactly when $$ 1 \lt |x - 2| \lt 5. $$ This inequality can now be solved through experimentation or by writing down more inequalities. For example when $x \geq 2$ we have $|x - 2| = x - 2$ and our inequalities become $1 \lt x - 2 \lt 5$. This has $x = 4, 5, 6$ as solutions. When $x \lt 2$ then the inequalities become $1 \lt -x + 2 \lt 5$. This has $x = 0, -1, -2$ for solutions. Thus the possible values of $x$ are -2,-1,0,4,5, and 6. The sum of these numbers is 12.

Solution: 3 Change of variables

Suppose we set $y =x - 2$. Then with this new variable our inequality becomes $$ 1 \lt y^2 \lt 25. $$ The only perfect squares in the required range are 4, 9, and 16 corresponding to $y$ values of $\pm$2, $\pm$3, and $\pm$4. Because the $\pm y$ solutions cancel each other out when added, the sum of the $y$ solutions is 0. Each $x$ solution is obtained from the corresponding $y$ solution by adding 2. Since there are 6 $y$ solutions, this means that the sum of the $x$ solutions will be 6 $\times$ 2 = 12.