Desmos for the SAT · Technique 16 of 23

Find the Constant That Gives One Solution, or None

Drag the constant until the graph shows one crossing, or two, or none.

SolvesNumber of solutionsOne, two, or no solutionsTangencyLine and parabola

Watch the lesson

The whole move worked on one real SAT question, narrated.

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A second equation, x^2 + 6x + c = 0, worked the same way: put c on a slider and move it until the two crossings become one.

Some of the trickiest SAT questions hand you an equation with an unknown constant and ask for the value that makes it have exactly one solution, or two, or none, or infinitely many.

Count the crossings with the x-axis, not the gray dots: at the tangent value Desmos marks two dots, the touch point and the y-intercept, and only one of them is a solution. What each picture means, and how many meetings a given pair of shapes can have at all, is the reference table in the systems lesson; this lesson works in reverse, from a count to the constant that produces it.

The by-hand way is the discriminant, b squared minus 4ac, and a sign or a factor often goes wrong there. The graph shows you the answer before any of that algebra.

Type the equation with the constant as a letter, turn that letter into a slider, and drag it. Watch how many times the graph crosses the x-axis. The value where you get the count the question wants is your answer. If the slider itself is new to you, the sliders lesson covers the mechanic; here it is only the tool.

Nova

Here's a whole family of hard questions that turn easy the second you see them as a graph: the ones asking for the constant that gives 'exactly one solution' or 'no solution.' Instead of doing the discriminant algebra, you drag a slider and literally watch the crossings appear and disappear.

Now you try it

The rest of the page takes that move slowly. First how it compares with doing the algebra, then the steps in order, then a real SAT question with a live Desmos calculator to try it in.

The algebra way vs the Desmos way

This technique sidesteps setting the discriminant $b^{2} - 4ac$ to zero and solving.

The question

$x^{2} + 8x + c = 0$ , with exactly one real solution

What is the value of c?

The algebra way

Write the discriminant: b^2 - 4ac = 8^2 - 4(1)(c).
Set it to zero for one solution: 64 - 4c = 0.
Solve: c = 16. Drop the 4 by accident and you get c = 64, which is exactly the trap answer.

The Desmos way

Type y = x^2 + 8x + c and add a slider for c.
Drag c until the parabola touches the x-axis at one point, landing on c = 16.

The payoff: The slider skips the $b^{2} - 4ac$ algebra entirely, which is exactly where the "drop the 4" error lives and where the wrong-answer choices are hiding.

Step by step

The whole move, all 5 steps

Type the equation as y = (the expression), using a letter for the unknown constant.
Click “add slider” when Desmos offers it.
Drag the slider and count the crossings on the x-axis: two crossings is two solutions, one touch is one, clear of the axis is none.
Stop at the value where the count matches what the question asked for.
Read the constant off the slider.

The question

In the equation $x^{2} + 8x + c = 0$ , c is a constant, and the equation has exactly one real solution.

What is the value of c?

Step 1 of 5

Type the equation as y = (the expression), using a letter for the unknown constant, e.g. y = x² + 8x + c.

In the calculator: y=x^2+8x+c

Set it up as y equals the left side. The solutions to the original equation are just wherever this graph hits the x-axis.

Now try it on a real SAT question

Head or scratch paper. Look at the coefficients before you reach for a slider. 8x is four times 2x and 28y is four times 7y, so the second equation is the first one multiplied by 4, and a has to be four times 9. That is the fastest route here, and it is exact.

In the system of equations $2x+7y=9$ and $8x+28y=a$ , a is a constant.

If the system has infinitely many solutions, what is the value of a?

Pick an answer to get instant coaching from Nova.

Stuck?

Your turnstarts empty

Set this question up yourself, then read the answer straight off the graph.

Worked it out on scratch paper? Type it into Desmos anyway and check. It takes a couple of seconds and it catches the slip you cannot spot by re-reading your own work.

The Desmos trap: when to close the calculator

A slider lands near a value, not on it. When the count changes at a fraction the slider cannot show cleanly, the discriminant gives the exact value; the drag tells you which way to go.
A slider often reveals two values that both work, like $k = 4$ and $k = -4$ . Before you answer, reread the problem for a constraint like k > 0 and pick the value that fits.

If the calculator does something you did not expect

Sliders default to a range of -10 to 10, so an SAT-scale constant like 25 or 50 sits off the end where you cannot reach it. Click the small numbers at the ends of the slider and widen the range.

All Desmos techniques

Put this move into real practice

"Find the Constant That Gives One Solution, or None" shows up all over the real SAT. SAT Prep Quest drills the skill at your level, so reaching for it becomes automatic by test day.

Practice free

SAT is a registered trademark of the College Board, which was not involved in producing, and does not endorse, this product. Score estimates are Learner Labs projections, not official College Board scores.