Two Squares are Equal

Task

Solve the quadratic equation $$ x^2 = (2x - 9)^2 $$ using as many different methods as possible.

IM Commentary

This classroom task is meant to elicit a variety of different methods of solving a quadratic equation (A-REI.4). Some are straightforward (for example, expanding the square on the right and rearranging the equation so that we can use the quadratic formula); some are simple but clever (reasoning from the fact that x and (2x - 9) have the same square); some use tools (using a graphing calculator to graph the functions $f(x) = x^2$ and $g(x) = (2x - 9)^2$ and looking for values of x at which the two functions intersect). Some solution methods will work on an arbitrary quadratic equation, while others (such as the last three) may have difficulty or fail if the quadratic equation is not given in a particular form, or if the solutions are not rational numbers.

It is recommended that teachers allow students the opportunity to discuss this problem in small groups of three or four students each. The small-group discussion will help students generate more solutions and reflect more carefully on the reasoning in each. Depending on how the class discussion develops, this task can touch on A-REI.4.a, A-REI.4.b, and A-REI.11.

In each solution below, we have taken care to show all of our reasoning about the equation and the solution process. The standards in domain A-REI ask students not only to carry out a procedure for solving a quadratic equation, but to understand such a procedure as a logical process in which an equality or a set of possible equalities at each step follows from the previous steps. Teachers should model this reasoning process when discussing problems with students.

Solutions

Solution: Method 1

We have $$ x^2 = (2x - 9)^2. $$ Expanding the right side yields $x^2 = 4x^2 - 36x + 81$. The two sides of this equation will still be equal if we subtract $x^2$ from each; therefore, we have $3x^2 - 36x + 81 = 0$. We can factor out a 3 to get $3(x^2 - 12x + 27) = 0$. Since the product of 3 and $x^2 - 12x + 27$ is zero, one of the factors must be zero; this means that $x^2 - 12x + 27 = 0$.

We can factor $x^2 - 12x + 27$ to obtain the equation $$ (x - 3)(x - 9) = 0. $$ Again, we have a product of two factors that is equal to zero; thus one of the two factors must be zero. So either $x - 3 = 0$ or $x - 9 = 0$. If $x - 3 = 0$, then $x = 3$. If $x - 9 = 0$, then $x = 9$. So the only possible solutions are $x = 3$ and $x = 9$. Substituting both of these into the original equation confirms that both are indeed solutions. Therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 2

We begin the same way we started in Method 1, expanding the right side and rearranging terms to obtain $3x^2 - 36x + 81 = 0$. This time, we subtract 81 from each side to find that $3x^2 - 36x = -81$. Since the two sides of the equation are equal, they will remain equal if we divide each side by 3. So we have $x^2 - 12x = -27$.

We will now solve the equation by completing the square. We recognize that, when expanded, the perfect square $(x - 6)^2$ contains the terms $x^2$ and $-12x$; it also contains a constant term of $36$. So we will transform the left side into a perfect square by adding 36 to each side of the equation: $$ x^2 - 12x + 36 = -27 + 36 \qquad \Rightarrow \qquad (x - 6)^2 = 9 $$ Since the square of $(x - 6)$ is 9, $(x - 6)$ must be 3 or -3. If $x - 6 = 3$, then $x = 9$. If $x - 6 = -3$, then $x = 3$. So $x$ must be either 9 or 3. We can substitute these numbers into the original equation to confirm that they are both solutions. Therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 3

Again, we start the same way we began Method 1, expanding the right side and rearranging terms to obtain the equation $$ 3x^2 - 36x + 81 = 0. $$ Since we have written the equation in the form $ax^2 + bx + c = 0$ with $a = 3$, $b = -36$, and $c = 81$, we may now apply the quadratic formula to see that

Therefore, the possible values of $x$ are 9 and 3. Substituting these values into the original equation confirms that they are both solutions; therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 4

Since we have a square on each side of the equation, we know that we can rearrange the equation to produce a difference of two squares on one side. This is advantageous because we can factor an expression of the form $a^2 - b^2$ into $(a - b)(a + b)$.

Since the two sides of the given equation are equal, we know that they will remain equal if we subtract $x^2$ from each side: $$ (2x - 9)^2 - x^2 = 0 $$ We can now factor the left side to obtain $((2x - 9) - x)((2x - 9) + x) = 0$; that is, $(x - 9)(3x - 9) = 0$.

We now have a product of two factors that is equal to zero; therefore, one of the two factors must be zero. If $x - 9 = 0$, then $x = 9$. If $3x - 9 = 0$, then $3x = 9$, and thus $x = 3$. Thus the only possible values of $x$ are 9 and 3. We can substitute both of these into the original equation to see that they are indeed solutions. Therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 5

If two real numbers have the same square, then they must either be the same number or be opposites of each other. Therefore, we must have either $x = 2x - 9$ or $x = -(2x - 9)$.

If $x = 2x - 9$, then $x - 9 = 0$, and thus $x = 9$.

If $x = -(2x - 9)$, then $x = 9 - 2x$. So $3x = 9$, and thus $x = 3$.

Thus the only possible solutions to the equation are $x = 9$ and $x = 3$. We can use the original equation to check that each of these is indeed a solution. Therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 6

Let $f(x) = x^2$ and let $g(x) = (2x - 9)^2$. We are looking for values of $x$ at which the functions $f(x)$ and $g(x)$ have the same value. Therefore, we will try graphing the two functions on the same coordinate plane:



We are looking for values of $x$ where $f(x)$ and $g(x)$ are the same; this is equivalent to looking for values of $x$ at which the graph of $f(x)$ and the graph of $g(x)$ have the same height. In other words, we are looking for where the two graphs intersect. From the graph, we see that the two intersection points are (3, 9) and (9, 81).

The fact that the two graphs intersect at (3, 9) means that when $x = 3$, both $f(x)$ and $g(x)$ are equal to 9. Thus $x^2$ and $(2x - 9)^2$ are equal when $x = 3$, and thus $x = 3$ is a solution to the equation $x^2 = (2x - 9)^2$. Similarly, the intersection point at (9, 81) means that when $x = 9$, both $x^2$ and $(2x - 9)^2$ are equal to 81. So $x = 9$ is also a solution.

In order to see that these are the only two solutions, we must convince ourselves that the two intersection points we see are the only intersection points of the two graphs. One way to see this is to observe that each graph is a parabola, and the parabola $g(x) = (2x - 9)^2$ is steeper than the parabola $f(x) = x^2$ for values of x outside the viewing window.

Therefore, the solution set is $\lbrace 3, 9 \rbrace$.

Solution: Method 7

We want to find values of $x$ for which $x^2 = (2x - 9)^2$; that is, for which $(2x - 9)^2 - x^2 = 0$. Therefore, we define $h(x) = (2x - 9)^2 - x^2$, and look for values of $x$ at which $h(x) = 0$.

We graph the function $h(x)$:



To determine when $h(x) = 0$, we need to find values of $x$ at which the height of the graph of $h(x)$ is zero; that is, at which the graph crosses the $x$-axis. This appears to occur at $x = 3$ and at $x = 9$. By substituting both of these values into the original equation, we find that both are indeed solutions. Because the function $h(x)$ is quadratic, we know that its graph is a parabola and can cross the x-axis at most twice. Therefore, we know that $x = 3$ and $x = 9$ are the only solutions. The solution set, then, is $\lbrace 3, 9 \rbrace$.

Solution: Testing code, please disregard this solution

Let $f(x) = x^2$ and let $g(x) = (2x - 9)^2$. We are looking for values of $x$ at which the functions $f(x)$ and $g(x)$ have the same value. Therefore, we will try graphing the two functions on the same coordinate plane:



We are looking for values of $x$ where $f(x)$ and $g(x)$ are the same; this is equivalent to looking for values of $x$ at which the graph of $f(x)$ and the graph of $g(x)$ have the same height. In other words, we are looking for where the two graphs intersect. From the graph, we see that the two intersection points are (3, 9) and (9, 81).

The fact that the two graphs intersect at (3, 9) means that when $x = 3$, both $f(x)$ and $g(x)$ are equal to 9. Thus $x^2$ and $(2x - 9)^2$ are equal when $x = 3$, and thus $x = 3$ is a solution to the equation $x^2 = (2x - 9)^2$. Similarly, the intersection point at (9, 81) means that when $x = 9$, both $x^2$ and $(2x - 9)^2$ are equal to 81. So $x = 9$ is also a solution.