The Discriminant and Number of Solutions

Introduction

Every quadratic equation has a built-in signal — the discriminant — that tells you exactly how many real solutions exist before you solve anything. On the SAT's Advanced Math section (≈35% of the math score), these questions appear in two flavors: compute the discriminant to classify solutions and find the unknown constant that forces exactly one solution. Mastering this single formula unlocks both.

Core Concept

Start from the quadratic formula for $ax^2 + bx + c = 0$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The expression under the radical, $b^2 - 4ac$ , is the discriminant $\Delta$ . It controls everything:

$\Delta = b^2 - 4ac$	Number of real solutions	Parabola behavior
$\Delta > 0$	Two distinct real solutions	Crosses the $x$ -axis at two points
$\Delta = 0$	One real solution (repeated root)	Touches the $x$ -axis at exactly one point (vertex on $x$ -axis)
$\Delta < 0$	No real solutions	Entire parabola above or below the $x$ -axis

Quick illustration: For $x^2 - 6x + 5 = 0$ , compute $\Delta = 36 - 20 = 16 > 0$ → two real solutions ( $x = 1$ and $x = 5$ ). No solving required beyond the discriminant check.

Intersections with horizontal lines: The same logic applies when a parabola meets $y = k$ . Set $f(x) = k$ , rearrange to $f(x) - k = 0$ , then evaluate the discriminant of that new equation.

Key Formulas & Rules

Discriminant (must be memorized — not on the reference sheet):

\Delta = b^2 - 4ac

Classification rules (memorize all three):

\Delta > 0 \implies \text{two distinct real solutions}

\Delta = 0 \implies \text{exactly one real solution (repeated root)}

\Delta < 0 \implies \text{no real solutions}

Finding the constant for exactly one solution: Set $b^2 - 4ac = 0$ and solve for the unknown. This is the most common SAT use of the discriminant.

Repeated root location: When $\Delta = 0$ , the one solution is $x = \dfrac{-b}{2a}$ , which is the $x$ -coordinate of the vertex.

The quadratic formula itself is not on the reference sheet — memorize it, or derive it via completing the square.

Worked Examples

Example 1

The equation $3x^2 - 6x + k = 0$ has exactly one real solution. What is the value of $k$ ?

A) $2$ B) $3$ C) $6$ D) $9$

Solution:

For exactly one real solution, set the discriminant equal to zero:

b^2 - 4ac = 0

Here $a = 3$ , $b = -6$ , $c = k$ :

(-6)^2 - 4(3)(k) = 0

36 - 12k = 0

k = 3

Verify: Substituting $k = 3$ gives $3x^2 - 6x + 3 = 0 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0$ . One solution: $x = 1$ . ✓

Why the other options fail:

A) $k = 2$ : Results from a division error — correctly reaching $12k = 36$ but dividing by $18$ instead of $12$ .
C) $k = 6$ : Results from using $2ac$ instead of $4ac$ in the formula: $36 - 6k = 0 \Rightarrow k = 6$ .
D) $k = 9$ : Results from ignoring $a$ in the $4ac$ term, treating it as $4c$ : $36 - 4k = 0 \Rightarrow k = 9$ .

Answer: B) $3$

Example 2

How many distinct real solutions does $2x^2 + 3x - 5 = 0$ have?

A) Zero B) Exactly one C) Exactly two D) More than two

Solution:

Identify $a = 2$ , $b = 3$ , $c = -5$ . Compute the discriminant:

\Delta = b^2 - 4ac = (3)^2 - 4(2)(-5) = 9 + 40 = 49

Since $\Delta = 49 > 0$ , the equation has two distinct real solutions.

As a check: $x = \dfrac{-3 \pm 7}{4}$ , giving $x = 1$ and $x = -\dfrac{5}{2}$ .

Verify $x = 1$ : $2(1)^2 + 3(1) - 5 = 2 + 3 - 5 = 0$ ✓

Why the other options fail:

A) Zero: Results from treating $c$ as $+5$ instead of $-5$ : $\Delta = 9 - 40 = -31 < 0$ — a sign error on the constant.
B) Exactly one: Results from correctly computing $\Delta = 49$ but misremembering the rule — $\Delta > 0$ means two solutions, not one.
D) More than two: A quadratic (degree 2) can have at most two solutions by the Fundamental Theorem of Algebra.

Desmos check

Desmos check: Type 2x^2+3x-5 in Desmos and count the $x$ -intercepts — the graph crosses the $x$ -axis at two distinct points, confirming two real solutions.

Answer: C) Exactly two

Common Mistakes & Traps

Sign error on $c$ : If the equation is $ax^2 + bx - c = 0$ , then $c$ is negative. Forgetting this when computing $4ac$ is the single most common discriminant error.
Dropping $a$ from $4ac$ : Students often compute $b^2 - 4c$ instead of $b^2 - 4ac$ when $a \neq 1$ . Every term in $4ac$ matters.
Confusing $\Delta = 0$ with $\Delta > 0$ : The "one solution" case is $\Delta = 0$ , not $\Delta > 0$ . The rule " $\Delta > 0$ means two solutions" trips up students who associate "positive discriminant" with "one real solution."
Solving for the root instead of the constant: When the SAT asks for the value of a constant $k$ that gives exactly one solution, set $\Delta = 0$ and solve for $k$ — do not find the root of the quadratic (that's a separate step).
Forgetting to rearrange before applying the discriminant: If the equation is $x^2 - 4x + 7 = 3$ , you must first rewrite it as $x^2 - 4x + 4 = 0$ before reading off $a$ , $b$ , $c$ .
Confusing tangency with crossing: " $\Delta = 0$ " means the parabola is tangent to the line — it touches at exactly one point without crossing through. " $\Delta > 0$ " means the parabola crosses the line at two points.

Practice Questions

Question 1 (Student-produced response)

For what value of $k$ does the equation $kx^2 - 8x + 4 = 0$ have exactly one real solution? (Assume $k \neq 0$ .)

Show answer

Answer: $k = 4$

Set the discriminant equal to zero with $a = k$ , $b = -8$ , $c = 4$ :

(-8)^2 - 4(k)(4) = 0

64 - 16k = 0

k = 4

Verify: $4x^2 - 8x + 4 = 0 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1$ . One solution. ✓

Question 2

The graph of $y = x^2 + bx + 9$ is tangent to the $x$ -axis (touches it at exactly one point). Which of the following could be the value of $b$ ?

A) $3$ B) $6$ C) $9$ D) $36$

Show answer

Answer: B) $6$

"Tangent to the $x$ -axis" means $\Delta = 0$ . With $a = 1$ , $c = 9$ :

b^2 - 4(1)(9) = 0 \implies b^2 = 36 \implies b = \pm 6

$b = 6$ appears in the options.

Why the other options fail:

A) $3$ : $\Delta = 9 - 36 = -27 < 0$ → no $x$ -intercepts at all.
C) $9$ : Confuses $b$ with $c$ ; $\Delta = 81 - 36 = 45 > 0$ → two intercepts.
D) $36$ : Results from writing $b^2 = 36$ and then setting $b = 36$ (not taking the square root).

Desmos check

Desmos check: Type y = x^2 + 6x + 9 and observe the vertex sits exactly on the $x$ -axis at $(-3, 0)$ .

Question 3

How many times does the parabola $y = x^2 - 4x + 7$ intersect the line $y = 3$ ?

A) Zero B) Exactly one C) Exactly two D) More than two

Show answer

Answer: B) Exactly one

Set the expressions equal and rearrange:

x^2 - 4x + 7 = 3 \implies x^2 - 4x + 4 = 0

Compute the discriminant with $a=1$ , $b=-4$ , $c=4$ :

\Delta = 16 - 16 = 0

$\Delta = 0$ means exactly one intersection point. The root is $x = 2$ ; check: $(2)^2 - 4(2) + 7 = 3$ ✓.

Why the other options fail:

A) Zero: Results from adding $3$ instead of subtracting: $x^2 - 4x + 10 = 0$ , $\Delta = 16 - 40 = -24 < 0$ (sign error when rearranging).
C) Exactly two: Results from computing $\Delta = 0$ but misapplying the rule (thinking $\Delta = 0$ means two solutions).
D) More than two: Impossible for an intersection of a parabola and a line (at most 2 points).

Question 4

Which of the following quadratic equations has no real solutions?

A) $x^2 - 5x + 6 = 0$ B) $x^2 - 4x + 4 = 0$ C) $x^2 - 3x + 4 = 0$ D) $x^2 + 2x - 3 = 0$

Show answer

Answer: C) $x^2 - 3x + 4 = 0$

Compute each discriminant:

A: $25 - 24 = 1 > 0$ → two solutions
B: $16 - 16 = 0$ → one solution
C: $9 - 16 = -7 < 0$ → no real solutions ✓
D: $4 + 12 = 16 > 0$ → two solutions

Question 5 (Student-produced response)

How many real solutions does the equation $x^2 + 10x + 25 = 0$ have?

Show answer

Answer: $1$

Discriminant: $\Delta = (10)^2 - 4(1)(25) = 100 - 100 = 0$ .

Since $\Delta = 0$ , there is exactly one real solution. The repeated root is $x = -5$ . Check: $(-5)^2 + 10(-5) + 25 = 25 - 50 + 25 = 0$ ✓.

Note: The equation factors as $(x + 5)^2 = 0$ , confirming a single repeated root.

Connections

Solving Quadratic Equations (prerequisite): The discriminant lives inside the quadratic formula. Once you know $\Delta > 0$ , you proceed to the full formula to find the two roots; if $\Delta = 0$ , the root is simply $x = -b/(2a)$ .
Systems of Linear and Nonlinear Equations (sibling subtopic): When you substitute a linear equation into a quadratic to find intersection points, the resulting equation's discriminant tells you how many intersections exist — a direct application of this note's method.
Solving Formulas for a Variable (sibling subtopic): Setting up $\Delta = 0$ and solving for an unknown constant $k$ is an instance of rearranging a formula, a skill reinforced in that note.
Vertex form and parabola geometry: Because $\Delta = 0$ forces the vertex onto the $x$ -axis (or the line $y = k$ in horizontal-line problems), strong discriminant fluency supports any SAT question about parabola transformations or minimum/maximum values.

Introduction

Core Concept

Key Formulas & Rules

Worked Examples

Example 1

Example 2

Common Mistakes & Traps

Practice Questions

Connections

Figures

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