There is little doubt that for many digital SAT math problems, graphing is the most expeditious approach when using Desmos for assistance. Each situation is a bit different, but there can be some drawbacks to using graphing for solving certain types of equations, and in some cases, you may find that you prefer regression. When seeking both solutions to a quadratic equation, graphing is generally the better choice, as you can typically enter the equation unaltered, then you can view the relevant coordinates of points on the resulting graph to get the answers you need. However, there are some types of problems involving solving quadratic equation that require you to match answers in fraction form, which coordinate values don't use, or in a compound form such as -3 + √11, necessitating the evaluation of each answer choice to find a match with the decimal value displayed for coordinates. It's also possible you'll need to do some arithmetic with one or both of the solution values, so you'd have to read them off the graph and type them in; if any such values have decimals that repeat or that have more digits than Desmos will display (Desmos will display three decimals by default, and four if you zoom in far enough), you could end up with an imprecise result. Regression doesn't suffer from any of these shortcomings, so though solving a quadratic equation for both solutions using regression is an advanced technique, you might want to master it.

As explained previously, a regression statement will find only one value for each regression parameter (variable with unknown value). In the large majority of cases, that is sufficient for enabling you to answer a digital SAT math question using a direct conversion of a given (or modeled) equation into regression form, even when there might be two solutions, as is often the case with quadratic equations and other types of equations as well.

The problem might have a restriction that results in only a single solution, and you can include that restriction in your regression statement; the problem might ask for any solution, so whichever one Desmos gives you will qualify; the problem could indicate that there is only a single solution, using wording such as "what is the solution," which tips you off.

However, in a small minority of situations, you will need to find two solutions to a quadratic equation in order to be certain you can answer the question. This can be straightforward, as when a problem asks for both solutions, and it can also happen when the problem gives a list of possible solutions and it asks you to select the one that is valid. Even though in the latter case you are only being asked for one solution, you have to know what both of them are; if you attempt a simple regression that just represents the quadratic equation as given, you'll only get one of the solutions, and it's a toss-up as to whether it will be the solution that's among the answer choices.

Here's an example where using straightforward regression will find one of the solutions, but it's the wrong one; the solution Desmos produces isn't in the list of answer choices:

Using regression to find the values of both solutions to a standard form quadratic equation

In order for a regression to produce two solutions for one unknown in an equation, each of the two solutions must be represented by a distinct regression parameter (variable). Therefore, we must craft an equation, or a system of equations, that includes both solutions. The technique we have created for this situation when it involves a quadratic equation is to represent the equation as a system of two equations: one for the sum of the solutions, and one for the product of the solutions.

You should know that in a standard form quadratic equation ax² + bx + c = 0, the formula -b / a gives the sum of the solutions, and the formula c / a gives the product of the solutions. Therefore, if we label the solutions as p and q, we can form this system of equations:

We can represent this system in a Desmos regression statement the same way we would with any system of equations: place the expressions on the left side of the equations into a list on the left side of the regression statement, and place the expressions on the right side of the equations into a list on the right side of the regression statement. Desmos will find the values of the regression parameters p and q that make both equations valid, and, thus, we will have both solutions to the quadratic equation:

An important note about crafting the regression: because both expressions on the right side of the equations in the system have a denominator of a, we can pull that out and make it the denominator of a fraction that has the list of numerators as its numerator; we don't need to repeat " / a" for both expressions in the list on the right-hand side. Here's the optimized form, then, of a regression statement that will find both solutions for a quadratic equation in standard form:

This clever technique lets you use Desmos, which does not do algebra, as a quadratic equation solver without resorting to the far more tedious, time-consuming, and error-prone technique of entering the quadratic formula (twice).

There's another regression-based approach you can use which might be appealing because it involves less memorization, though it requires more typing. A quadratic function can be uniquely specified by supplying three points because there are three constants/coefficients involved; this is analogous to the fact that two points determine a line, where a linear equation has two constants/coefficients, such as m and b in the slope-intercept form y = mx + b of a linear equation.

The factored form of a linear equation, y = a(x - p)(x - q), exposes the roots p and q, so if we can convert a quadratic equation into that form, we'll have our roots. We can't just set, say, a standard form quadratic equation equal to a factored form equation with a regression statement, because Desmos will just choose values that make the equation true for some value of x, but that doesn't mean those values will work for any other values of x. However, if we supply as many different values of x as are needed to uniquely specify the quadratic expression, that will restrict Desmos to finding the only values of the unknowns that make the two versions of the quadratic expression equivalent.

A standard form quadratic equation has three constants/coefficients, conventionally labeled a, b, and c, and a factored form quadratic has a, p, and q, where p and q are the roots. Similarly, a vertex form quadratic has a, h, and k. Though we'd need three points (solutions, or x-y value pairs) to completely specify a quadratic function, you'll notice that all three forms share the coefficient a, so we don't actually need to solve for that. As a consequence, we only need to solve for two unknowns, and that means we only need two input x-values and the corresponding output y-values or function values to implement a regression operation that converts from either a standard form or vertex form quadratic equation to a factored form quadratic, which will deliver the equation's roots to us.

To set up this form-conversion operation, we simply assign any two values to x, then we write a regression statement that sets our quadratic equal to a factored form quadratic; Desmos will find values of the unknown roots p and q that will make the two quadratic expressions equal for both values of x, and therefore equal for all values of x. Importantly, when we write the factored form expression, we simply repeat the value of the a coefficient from the known form of the quadratic expression, leaving Desmos with just the two unknowns p and q.

Here's the previous example solved using this sum/product method and the form-conversion method; this time, we see both solutions, so we can select the answer choice that represents a solution present among the answer choices:

Solving standard form quadratic equations with answer choices that include radicals

There are occasionally digital SAT math problems that require you to solve a quadratic equation, but the answer asks for an expression of the form h ± √r or some variation, where h represents the x-coordinate of the vertex or axis of symmetry of the parabola corresponding to the quadratic expression and √r represents the distance between the axis of symmetry and the roots:

These are the forms of solutions that would be produced by solving with completing the square; the quadratic formula will also produce solutions that can be manipulated to match these forms. But if you want to use Desmos, you can set up a system of equations using the sum/product technique described above, but you'll have to use a different representation of the two solutions. If one solution is h + √r and the other solution is h - √r, the sum of the solutions would be h + √r + h - √r = 2h, and the product of the solutions would be (h + √r)(h - √r) = h² - r, so the system of equations would be

The corresponding regression statement template would be

This will solve for h and r, allowing you to construct the two solutions h ± √r.

You can also use the form-conversion approach, representing the roots in the factored form expression as h + √r and h - √r. We show both methods below.