What It Targets
- Quadratic Equations
- Number of Solutions
- Systems of Equations
- Intersections
- Tangency Questions
- Advanced Math
The Core Idea
The SAT loves asking questions about:
- one solution
- two solutions
- no real solutions
- one intersection
- no intersections
- tangent to the x-axis
- touches exactly once
Most students immediately start solving for x.
This is usually a trap.
The SAT often doesn't care what the solutions are.
It only cares how many solutions exist.
When that happens:
Use the Discriminant.
The Discriminant Formula
For any quadratic:
ax² + bx + c = 0
The discriminant is:
b² - 4ac
You do NOT need to solve the equation.
Just evaluate:
b² - 4ac
and interpret the result.
The Three Cases You Must Memorize
Case 1: Exactly One Real Solution
When:
b² - 4ac = 0
Result:
The parabola touches the x-axis exactly once.
The equation has exactly one real solution.
Other SAT wording:
- tangent to the x-axis
- intersects exactly once
- one repeated root
- one real solution
Case 2: Two Distinct Real Solutions
When:
b² - 4ac > 0
Result:
The parabola crosses the x-axis twice.
The equation has two different real solutions.
Other SAT wording:
- two solutions
- two distinct roots
- intersects twice
Case 3: No Real Solutions
When:
b² - 4ac < 0
Result:
The parabola never touches the x-axis.
The equation has no real solutions.
Other SAT wording:
- no solutions
- no real roots
- no intersections
Visual Memory Trick
Positive Discriminant
b² - 4ac > 0
Parabola cuts through the x-axis twice.
Two solutions.
Zero Discriminant
b² - 4ac = 0
Parabola kisses the x-axis.
One solution.
Negative Discriminant
b² - 4ac < 0
Parabola misses the x-axis completely.
No real solutions.
How Do I Know This Is a Discriminant Question?
Look for phrases like:
Clue #1
- exactly one solution
- one real solution
- one intersection
- tangent
Immediately think:
b² - 4ac = 0
Clue #2
- no real solutions
- does not intersect
- no x-intercepts
Immediately think:
b² - 4ac < 0
Clue #3
- two solutions
- two distinct solutions
- intersects twice
Immediately think:
b² - 4ac > 0
Clue #4
The SAT asks about the number of solutions but never actually asks you to find x.
Huge clue.
Use the discriminant.
Guided Practice
Question 1
The equation
3x² - 6x + c = 0
has exactly one real solution.
What is the value of c?
Answer: 3
The Trick
Exactly one solution means:
b² - 4ac = 0
Substitute:
(-6)² - 4(3)(c) = 0
36 - 12c = 0
c = 3
Question 2
For what value of b does the equation
x² + bx + 25 = 0
have no real solutions?
(A) b = -12
(B) b = -8
(C) b = 10
(D) b = 14
Answer: B
The Trick
No real solutions means:
b² - 4ac < 0
Substitute:
b² - 4(1)(25) < 0
b² < 100
Only:
b = -8
satisfies the condition.
Guided Practice
Question 3
The system below has no real solutions.
y = 2x² - 4x + 7
y = 2x + k
What is the value of k?
Answer
Any value less than 2.5
The Trick
Set the equations equal.
2x² - 4x + 7 = 2x + k
Move everything to one side.
2x² - 6x + (7 - k) = 0
No real solutions means:
b² - 4ac < 0
Substitute:
(-6)² - 4(2)(7-k) < 0
36 - 56 + 8k < 0
8k < 20
k < 2.5
Drill
Do NOT solve for x.
Use only the discriminant.
1
How many real solutions does the equation
-2x² + 5x - 4 = 0
have?
2
The graph of
f(x) = x² + kx + 9
is tangent to the x-axis.
If k > 0, what is k?
3
In the equation
x² - 8x + (16 - k) = 0
for what values of k will there be two distinct real solutions?
4
The line
y = mx - 4
intersects the parabola
y = x²
exactly once.
If m > 0, what is m?
5
For what value of a will the equation
ax² + 12x + 4 = 0
have exactly one real solution?
6
Determine the number of real solutions for the system:
y = -x² + 3
y = 3x + 6
7
The equation
x² + 6x + m = 0
has two distinct real solutions.
What is the maximum integer value of m?
8
For what values of p will the equation
x² + px + 49 = 0
have exactly one real solution?
9
The parabola
y = x² - 10x + c
touches the x-axis exactly once.
What is c?
10
How many real solutions does the equation
4x² + 4x + 5 = 0
have?
Answer Key
1
0 solutions
5² - 4(-2)(-4)
25 - 32
-7
Negative.
2
6
k² - 36 = 0
k = ±6
Given k > 0:
k = 6
3
k > 0
64 - 4(16-k) > 0
4k > 0
k > 0
4
4
Set equal:
x² - mx + 4 = 0
Exactly one solution:
m² - 16 = 0
m = ±4
Given m > 0:
m = 4
5
9
144 - 16a = 0
a = 9
6
0 solutions
Set equal:
x² + 3x + 3 = 0
Discriminant:
9 - 12
-3
Negative.
7
8
36 - 4m > 0
m < 9
Largest integer:
8
8
p = 14 or p = -14
p² - 196 = 0
p² = 196
p = ±14
9
25
(-10)² - 4(1)(c) = 0
100 - 4c = 0
c = 25
10
0 solutions
4² - 4(4)(5)
16 - 80
-64
Negative.
SAT Shortcut Summary
Whenever you see:
- one solution
- tangent
- touches once
Think:
b² - 4ac = 0
Whenever you see:
- two solutions
- two intersections
- distinct roots
Think:
b² - 4ac > 0
Whenever you see:
- no real solutions
- no intersections
- no x-intercepts
Think:
b² - 4ac < 0
Do not solve the quadratic unless the SAT actually asks for the value of x.