Worked question
Exactly one solution, worked two ways
This is a hard Digital SAT system — the kind that appears late in the second module when the first went well. Most students who lose the mark here do not lose it on the algebra. They lose it on a single word in the question.
The question
y = x2 + bx + 10
y = 2x + 1
In the given system of equations, b is a positive constant. The system has exactly one solution. What is the value of b?
An original question, written to the published specification for the Advanced Math domain. Work it before reading on — the two routes below are only useful once you have committed to an answer.
First, translate the sentence
Nothing can start until exactly one solution has been converted into something you can act on. A solution to the system is a point lying on both graphs. One equation is a parabola and the other is a straight line, so the question is asking for the value of b that leaves the line touching the parabola at a single point rather than cutting through it at two, or missing it entirely.
That is the whole question. Everything after it is routine. Students who stall here almost never stall on the mechanics that follow — they stall because they went looking for a technique before deciding what was being asked.
Route one — the discriminant
Both equations give y, so set the right-hand sides equal and collect everything on one side:
x2 + bx + 10 = 2x + 1
x2 + (b − 2)x + 9 = 0
A quadratic has exactly one solution when its discriminant is zero, so with a = 1, the middle coefficient (b − 2) and c = 9:
(b − 2)2 − 4(1)(9) = 0
(b − 2)2 = 36
b − 2 = 6 or b − 2 = −6
b = 8 or b = −4
The question says b is positive, so b = 8.
Route two — Desmos
Type both equations into the built-in graphing calculator exactly as printed. Desmos will not recognise b, and will offer to add a slider for it. Accept.
Now drag the slider. The line holds still and the parabola slides; at b = 8 it settles so the line grazes the curve at one point. Widen the slider's range and drag the other way and the same thing happens again at b = −4. Read off the positive one.
Roughly twenty seconds, against a minute or so for the algebra, and — this is the part that matters more than the speed — the second answer appears on screen whether or not you remembered to look for it.
The trap, which is not the algebra
The line that decides this question is b is a positive constant. It is not scene-setting. It is there because the working produces two values and the question wants one, and it is the only thing in the question that tells you which.
The common failure is not getting the discriminant wrong. It is writing (b − 2)2 = 36, taking the square root of both sides as b − 2 = 6, and arriving at 8 without ever noticing that −4 existed. The answer is right. The method is not, and the same method meets a question next month where the wanted root is the negative one and quietly returns the wrong number.
So the check is worth making a habit: if you reach a single answer and never used the constraint the question gave you, go back and find the root you dropped. A constraint on a constant's sign or range is almost always a signpost that there were two.
Which route should you actually take?
| Situation | Faster route | Why |
|---|---|---|
| Tangency, intersections, “how many solutions” | Desmos | The question is already a picture; the slider shows every answer at once |
| Answer is an exact surd or fraction | Algebra | A graph can only be read approximately, and the options will be exact |
| The unknown must stay symbolic | Algebra | Nothing to plot until a number exists |
| Ugly coefficients, clean answer | Desmos | Arithmetic slips cost more marks than method choice ever does |
| You cannot see a route within fifteen seconds | Desmos | Plotting something is a better use of the next thirty seconds than staring |
This is the position taught in lessons rather than a neutral survey: above about 1400, where both routes work, the graph usually wins — not because it is quicker, though it often is, but because it fails more visibly. Algebra that goes wrong still produces a confident-looking number. A graph that goes wrong looks wrong.
How to recognise the type
Strip this question of its numbers and what remains is a pattern that recurs across the Advanced Math domain: a system containing one unknown constant, plus a statement about how many solutions there are. It appears as exactly one solution, as no real solutions, as two distinct real solutions, and occasionally hidden inside a word problem about a projectile that just reaches a height.
Every version is the same three moves. Substitute to get a single quadratic. Set the discriminant to zero, or make it positive or negative as the wording demands. Solve for the constant, then check the wording for which root survives. Once the pattern is recognised the question stops being hard, which is the actual reason these are worth working slowly one at a time rather than sitting another paper. There is more on that in going from 1400 to 1600, and the wider picture of the section in SAT Math tutoring.
Common questions
What does exactly one solution mean for a system on the SAT?
When one equation is a line and the other a parabola, exactly one solution means the line touches the curve at a single point rather than cutting through it — it is tangent. Algebraically, substituting one equation into the other produces a quadratic, and a quadratic has exactly one solution precisely when its discriminant is zero. Setting the discriminant to zero is the move the question is testing, and the phrase exactly one solution is the signal that it wants it. A system with two solutions gives a positive discriminant and one with no solution gives a negative one.
How do you find the value that makes a line tangent to a parabola?
Substitute one equation into the other, collect everything to the form ax squared plus bx plus c equals zero, then set b squared minus 4ac equal to zero and solve for the unknown constant. Because that equation is itself usually quadratic in the constant, expect two answers and read the question again to see which one is wanted. The same job can be done on the built-in graphing calculator by putting a slider on the constant and dragging until the line just grazes the curve.
Is Desmos allowed on the SAT, and when should I use it?
The built-in Desmos graphing calculator is available throughout the Math section of the Digital SAT, on every question. Use it wherever a question can be restated as where do these two graphs meet, which covers most systems, intersections, tangency questions and a good many word problems. Use algebra when the answer has to be an exact expression, when the numbers are irrational and a graph can only be read approximately, or when a variable has to stay symbolic. Above roughly 1400 the habit worth building is knowing both routes on the same question and choosing between them deliberately rather than by reflex.
Why does it matter that a constant is described as positive?
Because it is the question quietly telling you that more than one value satisfies the algebra. A phrase constraining the sign or the range of a constant almost always appears because the working produces two candidates and only one is wanted. The practical test is this: if you arrive at a single answer and never once used that phrase, you have probably lost a root somewhere — most often by taking only the positive square root when undoing a square.
Send the question that went wrong
Send Lewis a question you got stuck on and a photo of the working. The free trial lesson is spent on that question, worked both ways, so you leave knowing which route you should be taking under time.
Prefer email? Write to lewis@altairprep.com.
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