One number counts the real solutions — and finds the constant that fixes the count.
In the given equation, c is a constant. If the equation has exactly one real solution, what is the value of c?
x2 + 12x + c = 0
Exactly one real solution means b² − 4ac = 0, so 12² − 4(1)c = 0, 144 = 4c and c = 36. Check: x² + 12x + 36 = (x + 6)², whose only solution is x = −6 ✓ B stops at half of 12 — 6 is the number inside the square, not the constant term; A uses b instead of b² (12 − 4c = 0); D drops the 4 in 4ac (144 − c = 0). At c = 6 the discriminant is 144 − 24 = 120, so that equation has two solutions, not one.
1
Get the equation to = 0, then name a, b and c with their signs.
The discriminant only reads a quadratic written as ax² + bx + c = 0. Anything else has to be moved over first.
3x2 − 7x
+ 2 = 0
a = 3b = −7c = 2
b is negative seven
The minus in front of 7x belongs to b. It will get squared away in a moment,
but writing b = 7 in an item where b really is negative is the most common
slip in this topic — and it matters the instant the SAT asks you to solve
for b rather than for the count.
2
The discriminant is the piece under the square root.
x = [−b ± √b² − 4ac] / (2a). Everything about the number of solutions lives inside that root.
D = b2 − 4ac
D = (−7)2 − 4(3)(2)
b in brackets, 4ac in order
D = 49 − 24
(−7)² = +49; 4 · 3 · 2 = 24
D = 25
positive
D = 25 > 0 → two real solutions
And indeed x = 7 ± 56 gives x = 2 and x = 13. But the question rarely wants
those — it wants the count, and you already have it without solving.
Write (−7)², not −7². The brackets are what keep the minus attached to the 7;
without them a calculator reads it as −(7²) = −49 and every later line is wrong.
3
The sign of D is the whole answer.
Positive → two. Zero → exactly one. Negative → none. Nothing in between.
If b² − 4ac is
Real solutions
Why
positive
exactly two
± a real number gives two different results
zero
exactly one
± 0 gives the same result twice
negative
zero
no real number squares to a negative
SAT wordinghow many distinct real solutionsexactly one real solutionno real solutionshow many x-intercepts
"Distinct" is doing real work in the first phrase. When D = 0 the formula still
produces two answers — they are just the same number, so the count of
distinct solutions is one.
4
On the graph, D counts how often the parabola meets the x-axis.
A real solution of ax² + bx + c = 0 is an x-intercept of y = ax² + bx + c. Same fact, two languages.
The three curves are the same parabola lifted by 4 and then by 4 again: the
lowest point moves from (3, −4) to (3, 0) to (3, 4). Raising c raises the whole
graph, and the two x-intercepts slide together, merge, and vanish.
The vocabulary the SAT swaps in
"The graph of y = ax² + bx + c intersects the x-axis at exactly one point",
"has no x-intercepts", "the system has no real solutions" when a parabola meets
a line — all three are the discriminant question wearing a different coat. Set
the equation to zero, take b² − 4ac, read the sign.
5
"Exactly one real solution" is an equation: b² − 4ac = 0.
This is where the discriminant earns its keep. The count is given; the constant is the unknown.
x2 + 12x + c = 0
a = 1, b = 12, c = c
122 − 4(1)(c) = 0
one solution → D = 0
144 − 4c = 0
square first
c = 36
÷ 4 both sides
c = 36, and x² + 12x + 36 = (x + 6)²
The unknown can sit in the b slot just as easily, and then squaring it hands you
two answers — so read whether the stem says the constant is positive.
4x2 + cx + 25 = 0, c positive
a = 4, b = c, c-term = 25
c2 − 4(4)(25) = 0
one solution → D = 0
c2 = 400
4 · 4 · 25 = 400
c = 20
c = ±20, and the stem says positive
Check: 4x² + 20x + 25 = (2x + 5)², one solution at x = −52 ✓ The other root,
c = −20, gives (2x − 5)² — also one solution, which is exactly why the stem has
to tell you which sign it wants.
6
"Two" or "none" gives an inequality instead.
Same substitution, one character different — and the answer becomes a range, so the choices ask "which could be the value".
two x-intercepts
y = x² − 8x + c
(−8)² − 4(1)c > 0
64 > 4c, so c < 16
no real solutions
9x² + 12x + c = 0
12² − 4(9)c < 0
144 < 36c, so c > 4
Test one value from each range before you commit. c = 10 in the first gives
64 − 40 = 24 > 0 ✓; c = 5 in the second gives 144 − 180 = −36 < 0 ✓
Thirty seconds, and the calculator does the arithmetic.
The boundary value is always in the choices
For "two x-intercepts" with c < 16, the number 16 itself will be sitting
there. At c = 16 the parabola touches the axis at one point, so it has
one x-intercept, not two. Strict inequality means the boundary is
excluded — that single number is the distractor the item was built around.
7
If both sides have terms, collect them before you take D.
b and c are whatever is left after everything moves to one side — not what you can see in the original.
x2 + 5x + k = 3x + 4
exactly one real solution; find k
x2 + 2x + (k − 4) = 0
− 3x and − 4 both sides
22 − 4(1)(k − 4) = 0
now b = 2 and c = k − 4
4 − 4k + 16 = 0
distribute the −4
k = 5
20 = 4k
k = 5
Reading b = 5 off the original line instead of b = 2 is the trap. The 5x and the
3x have to meet first.
Desmos is on your screen — use it
Type y = x² + 12x + c and Desmos offers you a slider for c. Drag it and watch
the parabola rise: at c = 36 the curve kisses the x-axis, and above that it
lifts clear. Use the slider to confirm a value you found by hand, or to settle
which side of a boundary an inequality answer falls on — it is far faster than
re-checking the sign three times.
✓
Check: substitute the constant back and count.
A discriminant answer is checkable in one line — factor the result and see whether it really is a perfect square.
c = 36x² + 12x + 36 = (x + 6)²one solution, x = −6 ✓
c = 6 insteadD = 144 − 24 = 120two solutions — the distractor fails
c = 20 in 4x² + cx + 25(2x + 5)², D = 400 − 400 = 0one solution ✓
k = 5 in step 7x² + 2x + 1 = (x + 1)², x = −1left side 1 − 5 + 5 = 1, right side 1 ✓
c > 4 test9x² + 12x + 5 → D = −36no real solutions ✓
Remember the flow
Move to = 0 · name a, b, c · b² − 4ac · read the sign
Count given instead? Set D = 0, > 0 or < 0 and solve for the constant.
Now you try
Practice problem
In the xy-plane, the graph of the given equation, where c is a constant, does not intersect the x-axis. Which of the following could be the value of c?
y = 2x2 − 8x + c
No x-intercepts means b² − 4ac < 0: (−8)² − 4(2)c < 0, so 64 < 8c and c > 8. Of the choices only 10 qualifies, and 64 − 80 = −16 < 0 ✓ C is the boundary — at c = 8 the discriminant is exactly 0, so the parabola touches the axis at (2, 0) instead of missing it. A and B sit below the boundary: c = 2 gives D = 48 and c = 4 gives D = 32, both positive, so those parabolas cross the axis twice.
Hint: "does not intersect the x-axis" means no real solutions,
so the discriminant is strictly negative. Watch the a = 2 — it belongs inside
the 4ac — and then check whether the borderline choice satisfies the strict
inequality or only the equation.
