SAT Quadratic Equations Part 2: Hard Questions, Parabolas, Parameters, Word Problems & Shortcuts
Quadratic equations become much more interesting on the harder SAT questions.
You may be asked to find a root, interpret a parabola, determine a maximum value, identify a parameter, connect an equation to a graph, or solve a real-world problem.
The key is recognizing that these are not separate topics.
They are different ways of using the same quadratic relationship.
A quadratic equation generally has the form:
ax² + bx + c = 0
where:
a ≠ 0
The graph of a quadratic function has the shape of a parabola.
This guide focuses on the harder patterns you should recognize quickly.
1. The Three Forms of a Quadratic
A quadratic can appear in three especially useful forms.
Standard form
f(x) = ax² + bx + c
Useful for identifying:
• a
• b
• c
• y-intercept
Factored form
f(x) = a(x − r₁)(x − r₂)
Useful for finding:
• roots
• zeros
• x-intercepts
Vertex form
f(x) = a(x − h)² + k
Useful for finding:
• vertex
• maximum or minimum
• axis of symmetry
Learning to move between these forms is one of the most useful SAT quadratic skills.
2. Roots and Zeros
Suppose:
f(x) = (x − 3)(x + 5)
The zeros occur when:
f(x) = 0
Therefore:
(x − 3)(x + 5) = 0
So:
x − 3 = 0
or:
x + 5 = 0
Therefore:
x = 3 or x = −5
These values are called:
• roots
• zeros
• solutions
They all refer to the x-values where the quadratic equals zero.
3. X-Intercepts
The x-intercepts of a parabola occur where:
y = 0
Therefore, finding the x-intercepts means solving:
f(x) = 0
For:
y = (x − 2)(x + 6)
the x-intercepts are:
(2, 0)
and:
(−6, 0)
The roots are:
2 and −6
4. The Vertex
The vertex is the highest or lowest point of a parabola.
If:
f(x) = a(x − h)² + k
the vertex is:
(h, k)
For example:
f(x) = (x − 4)² + 7
the vertex is:
(4, 7)
Because the coefficient of the squared term is positive, the parabola opens upward.
Therefore, the vertex represents a:
minimum
5. When the Parabola Opens Downward
Consider:
f(x) = −(x − 3)² + 10
The vertex is:
(3, 10)
The leading coefficient is negative.
Therefore, the parabola opens downward.
The vertex is therefore the:
maximum
value.
Quick rule
a > 0 → minimum
a < 0 → maximum
6. Finding the Vertex From Standard Form
Suppose:
f(x) = x² − 8x + 11
The x-coordinate of the vertex is:
x = −b/(2a)
Here:
a = 1
b = −8
Therefore:
x = −(−8)/(2)
x = 4
Now substitute x = 4:
f(4) = 16 − 32 + 11
f(4) = −5
Therefore, the vertex is:
(4, −5)
7. The Axis of Symmetry
The axis of symmetry passes through the vertex.
For:
f(x) = ax² + bx + c
the axis is:
x = −b/(2a)
For:
f(x) = 2x² − 12x + 7
we have:
a = 2
b = −12
Therefore:
x = 12/4
x = 3
Axis of symmetry:
x = 3
8. A Useful Symmetry Trick
Suppose the roots of a quadratic are:
2 and 10.
The axis of symmetry lies exactly halfway between them.
Therefore:
(2 + 10)/2 = 6
Axis of symmetry:
x = 6
This can be faster than using:
−b/(2a)
when the roots are already known.
9. The Vertex From Two Roots
Suppose:
f(x) = (x − 2)(x − 10)
The roots are:
2 and 10.
Therefore, the axis of symmetry is:
x = 6
Now substitute:
f(6) = (6 − 2)(6 − 10)
= 4(−4)
= −16
Therefore:
Vertex = (6, −16)
10. Completing the Square
Completing the square is useful for converting standard form into vertex form.
Consider:
x² + 6x + 5
Take half of 6:
3
Square it:
9
Rewrite:
x² + 6x + 9 − 9 + 5
Group:
(x + 3)² − 4
Therefore:
x² + 6x + 5 = (x + 3)² − 4
The vertex is:
(−3, −4)
11. Completing the Square With a Leading Coefficient
Consider:
2x² + 12x + 7
First factor 2 from the quadratic terms:
2(x² + 6x) + 7
Complete the square:
2[(x + 3)² − 9] + 7
Expand:
2(x + 3)² − 18 + 7
Therefore:
2(x + 3)² − 11
The vertex is:
(−3, −11)
12. Quadratic Formula
When factoring is difficult, use the quadratic formula.
For:
ax² + bx + c = 0
the solutions are:
x = [−b ± √(b² − 4ac)]/(2a)
For example:
x² − 5x + 6 = 0
Here:
a = 1
b = −5
c = 6
Therefore:
x = [5 ± √(25 − 24)]/2
x = [5 ± 1]/2
So:
x = 3
or:
x = 2
13. The Discriminant
The expression:
b² − 4ac
is called the discriminant.
It tells you how many real solutions the quadratic has.
If:
b² − 4ac > 0
There are:
two distinct real solutions
If:
b² − 4ac = 0
There is:
one real solution
If:
b² − 4ac < 0
There are:
no real solutions
This connects algebra directly to the graph.
14. Discriminant and the Graph
If a quadratic has two real roots, its parabola crosses the x-axis twice.
If it has one real root, the parabola touches the x-axis once.
If it has no real roots, the parabola does not cross the x-axis.
Therefore:
Two roots
→ Two x-intercepts
One root
→ One x-intercept
No real roots
→ No x-intercepts
This is a powerful connection between algebra and graphs.
15. Parameter Questions
Hard SAT questions may include a variable parameter.
For example:
x² − 6x + k = 0
For what value of k does the equation have exactly one real solution?
Exactly one real solution means:
discriminant = 0
Therefore:
b² − 4ac = 0
Here:
a = 1
b = −6
c = k
So:
36 − 4k = 0
4k = 36
k = 9
Answer:
9
16. Parameter Question: Two Real Solutions
Consider:
x² − 4x + k = 0
For two distinct real solutions:
b² − 4ac > 0
Therefore:
16 − 4k > 0
−4k > −16
k < 4
Answer:
k < 4
This is an example where you are not looking for one number.
You are looking for a range of values.
17. Parameter Question: No Real Solutions
Consider:
x² + 2x + k = 0
For no real solutions:
b² − 4ac < 0
Therefore:
4 − 4k < 0
−4k < −4
k > 1
Answer:
k > 1
18. Quadratic Equations From Graphs
Suppose a graph crosses the x-axis at:
x = −2
and:
x = 5
Then the quadratic can be written as:
f(x) = a(x + 2)(x − 5)
If the graph also passes through a known point, you can determine a.
Suppose the graph passes through:
(0, 20)
Then:
20 = a(2)(−5)
20 = −10a
a = −2
Therefore:
f(x) = −2(x + 2)(x − 5)
19. Finding a Quadratic From Three Points
A quadratic can be written:
y = ax² + bx + c
If three points are given, substitute each point to create three equations.
For example, suppose:
(0, 2)
(1, 6)
(2, 12)
From (0, 2):
c = 2
From (1, 6):
a + b + 2 = 6
a + b = 4
From (2, 12):
4a + 2b + 2 = 12
4a + 2b = 10
Solving gives:
a = 1
b = 3
c = 2
Therefore:
y = x² + 3x + 2
20. Quadratic Word Problems
Many SAT quadratic word problems involve area.
Suppose a rectangle has length:
x + 5
and width:
x
Its area is:
x(x + 5)
Therefore:
A = x² + 5x
If the area is 36:
x² + 5x = 36
or:
x² + 5x − 36 = 0
Factor:
(x + 9)(x − 4) = 0
Therefore:
x = −9
or:
x = 4
Since a length cannot be negative:
x = 4
21. Why Negative Roots Can Be Rejected
Algebraically, x = −9 is a solution to the equation.
But if x represents a physical length, negative 9 is impossible.
Therefore:
x = 4
This is a common SAT skill.
The equation may produce multiple mathematical solutions, but the context may allow only one.
22. Area Problems
Suppose a rectangle has dimensions:
x + 2
and:
x + 5
Its area is 48.
Then:
(x + 2)(x + 5) = 48
Expand:
x² + 7x + 10 = 48
x² + 7x − 38 = 0
At this point, you can use factoring if possible or the quadratic formula.
The important step is translating the geometry into an equation.
23. Projectile Problems
Quadratics frequently model height.
A typical model might be:
h(t) = −16t² + 64t + 5
where:
t is time and:
h(t) is height.
The negative coefficient tells you the parabola opens downward.
Therefore, the object eventually reaches a maximum height.
24. Finding Maximum Height
For:
h(t) = −16t² + 64t + 5
the time at maximum height is:
t = −b/(2a)
Therefore:
t = −64/[2(−16)]
t = 2
Now calculate:
h(2) = −16(4) + 64(2) + 5
= −64 + 128 + 5
= 69
Maximum height:
69
25. Finding When an Object Hits the Ground
If height is zero, set:
h(t) = 0
For:
h(t) = −16t² + 64t + 5
solve:
−16t² + 64t + 5 = 0
The resulting solutions represent times when the height is zero.
A negative time would normally be rejected because the physical situation begins at:
t = 0
26. Maximum and Minimum Questions
Whenever the SAT asks:
• greatest value
• maximum
• highest point
• minimum
• lowest value
and the equation is quadratic, immediately think:
vertex
For:
f(x) = ax² + bx + c
find:
x = −b/(2a)
Then substitute to find the corresponding y-value.
27. Vertex Form Makes Maximum Questions Easy
Suppose:
f(x) = −2(x − 4)² + 13
The vertex is:
(4, 13)
Because the parabola opens downward, the maximum value is:
13
No calculation is required.
28. Comparing Two Quadratics
Suppose:
f(x) = x²
and:
g(x) = x² − 6x + 8
To find where they are equal:
x² = x² − 6x + 8
Subtract x²:
0 = −6x + 8
6x = 8
x = 4/3
This is another example of finding an intersection between two functions.
29. Quadratic vs. Linear
Suppose:
f(x) = x² − 4x + 3
and:
g(x) = x + 3
To find where they intersect:
x² − 4x + 3 = x + 3
x² − 5x = 0
x(x − 5) = 0
Therefore:
x = 0
or:
x = 5
There are two intersection points.
30. A Common SAT Graph Question
A parabola crosses the x-axis at:
−3 and 7.
Without doing any algebra, what is the axis of symmetry?
Take the midpoint:
(−3 + 7)/2
= 4/2
= 2
Answer:
x = 2
This is much faster than deriving the entire equation.
31. Finding the Constant From a Root
Suppose:
f(x) = x² + kx − 12
and x = 3 is a root.
Then:
f(3) = 0
Substitute:
9 + 3k − 12 = 0
3k − 3 = 0
k = 1
Answer:
1
Whenever the question says a particular value is a solution or root, substitute it directly.
32. Finding a Parameter From a Point
Suppose:
f(x) = x² + kx + 4
and the graph passes through:
(2, 10)
Therefore:
f(2) = 10
4 + 2k + 4 = 10
2k + 8 = 10
2k = 2
k = 1
Answer:
1
The same principle works for many function questions.
33. Root Relationships
If:
ax² + bx + c = 0
has roots:
r₁ and r₂
then:
r₁ + r₂ = −b/a
and:
r₁r₂ = c/a
These relationships can save time.
34. Example Using Root Sum
Suppose the roots of:
2x² − 10x + 7 = 0
are r₁ and r₂.
What is:
r₁ + r₂?
Use:
r₁ + r₂ = −b/a
Therefore:
10/2 = 5
Answer:
5
You do not need to solve the quadratic.
35. Example Using Root Product
For:
3x² + 8x − 10 = 0
the product of the roots is:
c/a
Therefore:
−10/3
Answer:
−10/3
Again, there is no need to use the quadratic formula.
36. A Powerful SAT Shortcut
If the question asks for:
sum of roots
→ use:
−b/a
If it asks for:
product of roots
→ use:
c/a
Do not solve the equation unless the actual roots are required.
37. Transformations of Quadratics
Suppose:
f(x) = x²
Then:
f(x) + 5
moves the graph:
5 units upward
while:
f(x) − 3
moves it:
3 units downward
Similarly:
f(x − 4)
moves the graph:
4 units right
and:
f(x + 2)
moves it:
2 units left
The signs inside the parentheses can be especially tricky.
38. The Horizontal Shift Trap
Compare:
f(x − 3)
and:
f(x + 3)
The first shifts right by 3.
The second shifts left by 3.
Remember:
Inside the parentheses, the direction appears reversed.
39. Effect of the Leading Coefficient
In:
f(x) = ax²
the value of |a| affects the width.
If:
|a| > 1
the parabola is narrower.
If:
0 < |a| < 1
the parabola is wider.
If:
a < 0
the parabola opens downward.
40. A Difficult Parameter Problem
Consider:
f(x) = x² − 8x + k
For the graph to touch the x-axis exactly once, what must k equal?
Touching the x-axis once means:
one real root
Therefore:
discriminant = 0
b² − 4ac = 0
64 − 4k = 0
k = 16
Answer:
16
41. A Harder Interpretation
Suppose:
f(x) = x² − 8x + k
For the graph to have no x-intercepts:
64 − 4k < 0
Therefore:
k > 16
So:
k > 16
This means the parabola lies entirely above the x-axis because its leading coefficient is positive.
42. Another Hard Question
Suppose:
f(x) = −x² + 6x + k
For the maximum value of f(x) to equal 10, find k.
The vertex occurs at:
x = −6/[2(−1)]
= 3
Now substitute:
10 = −(3²) + 6(3) + k
10 = −9 + 18 + k
10 = 9 + k
k = 1
Answer:
1
43. The SAT “Which Equation?” Question
Sometimes you are shown a graph and several equations.
Look for:
Direction
Parabola opens upward:
a > 0
Parabola opens downward:
a < 0
Vertex
Compare the vertex coordinates.
X-intercepts
Check the roots.
Y-intercept
Set:
x = 0
The correct equation must match all the available information.
44. Use the Cheapest Information First
If the graph clearly shows:
• x-intercepts
• vertex
• y-intercept
do not calculate everything.
Use the easiest visible feature to eliminate incorrect answers.
For example, if the parabola opens downward, immediately eliminate every equation with a positive x² coefficient.
This is a powerful multiple-choice strategy.
45. SAT Quadratic Checklist
Before test day, make sure you can:
✓ Factor quadratics.
✓ Use the quadratic formula.
✓ Find roots.
✓ Find x-intercepts.
✓ Find the vertex.
✓ Find the axis of symmetry.
✓ Complete the square.
✓ Use the discriminant.
✓ Determine the number of real solutions.
✓ Solve parameter questions.
✓ Interpret quadratic graphs.
✓ Find maximum and minimum values.
✓ Solve quadratic word problems.
✓ Reject impossible contextual solutions.
✓ Use root sum and root product.
✓ Recognize transformations.
✓ Compare quadratic functions.
✓ Find a parameter using a known root or point.
Final SAT Quadratic Strategy
When you see a quadratic, do not automatically reach for the quadratic formula.
First ask:
What is the question actually asking?
If it asks for the roots:
→ factor or use the quadratic formula.
If it asks for the x-intercepts:
→ find the roots.
If it asks for the maximum or minimum:
→ find the vertex.
If it asks for the axis of symmetry:
→ use −b/(2a), or the midpoint of the roots.
If it asks how many real solutions exist:
→ use the discriminant.
If it gives a parameter and asks for exactly one solution:
→ set the discriminant equal to zero.
If it gives a known root:
→ substitute it directly.
If it asks for the sum of roots:
→ use −b/a.
If it asks for the product of roots:
→ use c/a.
If it gives a graph:
→ use the vertex, intercepts, direction, and symmetry.
The biggest SAT quadratic skill is therefore not memorizing more formulas.
It is recognizing which piece of the quadratic contains the answer.
Once you can identify that quickly, many of the hardest-looking quadratic questions become short problems.
