One problem, solved five different ways

I came across a math problem in a session with a student that I realized can be solved with a variety of different techniques. It is often useful as a tutor to have many different ways to solve the same problem on hand, as students may respond to the different techniques in various ways. The original problem was: 

y = 2x² - 21x + 64

y = 3x + d

In the given system of equations, d is a constant. The graphs of the equations in the given system intersects at exactly one point (x,y), in the xy-plane. What is the value of x?

We can solve that math problem in four different ways:

1. Substitution and the Quadratic Formula 

My initial thought when I first encountered this problem was to use substitution to set the equations equal to each other. Substitution, along with graphing and elimination, is one of the main ways to solve systems of equations problems, and it jumped out at me here because both equations were already equal to y. Setting them equal to each other, we get 

3x + d = 2x² - 21x + 64

We now have one quadratic equation, and the first step in solving any quadratic equation is to set the equation equal to 0:

0 = 2x² - 24x + 64 - d

The fact that there is only one solution tells me that the discriminant ∆ = b² - 4ac must be 0. Using a = 2, b = -24, and c = 64 -d, I can use the discriminant equation to solve for d. Note that, since c in the quadratic formula is the entire constant terms (in other words, anything not attached to an x), in this case d is part part of the constant term. Using the discriminant equation, I get 

0 = (-24)² - 4 · 2 · (64 - d)

0 = 576 - 512 + 8d

-64 = 8d

-8 = d

Now that I now the value of d, I can plug it back into the original equation to get 0 = 2x² - 24x + 72 and use the entire quadratic formula to solve for the final answer, x: 

Therefore, x = 6 is the final solution to the problem! (The original question only asked for the x-coordinate of the intersection, so we don’t have to plug x = 6 back in anywhere.) However, I soon realized that there was a simpler way to solve the problem.

2. Ignoring the discriminant

Since the problem has already told us that there is one solution, I already know the discriminant b² - 4ac has to be zero. That means, when I plug everything into the quadratic formula, I know I would get 

It turns out I didn’t have to find the value of d at all! Since d is only a part of the c term in the quadratic and, in the quadratic formula, c only appears in the discriminant (which I already know has to be zero, even without finding d). I can just plug zero in for b² - 4ac, leaving me with -b(2/a) and since I already know that b = -24 and a = 2, I can just plug those values in and solve the problem! 

3. Using the Vertex

If you have experience with quadratics, you may realize that the final answer -b(2/a) = 6 is just the formula for the vertex of the quadratic. This is because, when a quadratic has only one solution, that means that the quadratic is only touching the x-axis at exactly one point, and the only point at which that can happen is the vertex. Therefore, the solution must be the vertex! So, I could have just dispensed with the quadratic formula entirely, recognized that the one solution was occurring at the parabola’s vertex, and found the vertex of the parabola at -b(2/a).

4. Graphing

Now that the SAT is online, many tools are available to students on desmos to help solve problems quickly with the graphing utility. I can plug the original two equations into desmos where, when I type the d in the second equation, desmos will provide me with a slider:

Click add slider to make the second line appear, and then you can move the slider to adjust the value of d. Changing the value of d with the slider will slide the blue line up or down:

Since we know we want only one solution, we want the place where the lines intersect exactly one time. By playing around with the slider, I can find the value of d that makes that happen:

If I have one of the equations for the lines selected, desmos will highlight the intersection point, where I can again see that the lines intersect exactly once when d = -8 and the solution to the problem is the x-coordinate of the intersection, which is 6.

Bonus: Calculus

Calculus is not required for the SAT, but a cursory knowledge of derivatives can help solve problems like this quickly. Since there is only one solution, I know the lines must be tangent to each other. When two lines are tangent (touching exactly once), they must have the same slope at the point where they touch. I can easily see that slope of the second line is 3, but the slope of the quadratic is constantly changing! To the left of the vertex, the graph is going down, so the slope is negative; to the right of the vertex, the slope is positive; and the slope is zero exactly at the vertex. Here is where calculus comes in. I can use calculus to find the derivative of the quadratic, which will tell me its slope at any given point. The simplest way to find a derivative, denoted as y' is to use the power rule, which states that for any term axⁿ, the derivative of that term is n · ax^n-1. In other words, I drop the exponent down in front as a constant multiplier, and my new exponent is one less than the old exponent. I can do this for each term of the quadratic, yielding  

When we take a derivative, constant terms disappear (since the slope of a flat line is zero) and linear terms just drop the x (remember that the exponent on x in -21x is actually 1, but we just don't write it). Remember that y', the derivative, represents the slope of the quadratic equation, which is constantly changing.

Since the slope of the second line y = 3x + d is 3 and I know that the two lines will have the same slope at the point of tangency, I can plug 3 in for y' and solve for x to find where the quadratic has a slope of 3:

3 = 4x - 21

24 = 4x

6 = x

Again, I have found the final answer of z = 6, and here was another method where I didn’t actually have to find the value of d in order to solve the problem.

While any one of these five methods effectively solves the problem, some are faster than others, and some will also play better to different students’ strengths. As a tutor, having multiple ways to solve a problem at my disposal gives me a better chance of being able to teach the problem to each student in a way they’ll understand, and if a student is aware of multiple methods, they can pick the one that makes the best sense to them.