What is Nonlinear Equations and Functions?

Solve quadratic, absolute value, and exponential equations.

How can I study Nonlinear Equations and Functions effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 15 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Nonlinear Equations and Functions study guide free?

Yes — all study notes, flashcards, and practice problems for Nonlinear Equations and Functions on Study Mondo are free to access. No account is needed.

What course covers Nonlinear Equations and Functions?

Nonlinear Equations and Functions is part of the SAT Prep course on Study Mondo, specifically in the Passport to Advanced Math section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Nonlinear Equations and Functions?

Yes, this page includes 15 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Nonlinear Equations and Functions on the SAT

Beyond Linear: Types of Nonlinear Functions

Quadratic Functions

❓ Question:

Solve: $|x - 4| = 6$

Two cases:

Case 1: $x - 4 = 6 \implies x = 10$

Case 2: $x - 4 = -6 \implies x = -2$

Check: $|10 - 4| = |6| = 6$ ✓ and $|-2 - 4| = |-6| = 6$ ✓

Answer: $x = 10$ or $x = -2$

❓ Question:

Solve: $|x - 4| = 6$

Two cases:

Case 1: $x - 4 =$

❓ Question:

Solve: $|x - 4| = 6$

Two cases:

Case 1: $x - 4 =$

❓ Question:

Solve: $\sqrt{3x + 1} = 4$

❓ Question:

Solve: $\sqrt{3x + 1} = 4$

❓ Question:

Solve: $\sqrt{3x + 1} = 4$

❓ Question:

How many solutions does the system $y = x^2 - 3$ and $y = 2x - 1$ have?

❓ Question:

How many solutions does the system $y = x^2 - 3$ and $y = 2x - 1$ have?

❓ Question:

How many solutions does the system $y = x^2 - 3$ and $y = 2x - 1$ have?

❓ Question:

Solve: $\sqrt{x + 7} = x + 1$

❓ Question:

Solve: $\sqrt{x + 7} = x + 1$

❓ Question:

Solve: $\sqrt{x + 7} = x + 1$

❓ Question:

For what value of $k$ does the line $y = 3x + k$ intersect the parabola $y = x^2 + 2$ at exactly one point?

❓ Question:

For what value of $k$ does the line $y = 3x + k$ intersect the parabola $y = x^2 + 2$ at exactly one point?

❓ Question:

For what value of $k$ does the line $y = 3x + k$ intersect the parabola $y = x^2 + 2$ at exactly one point?

Step 1: Square both sides to eliminate the radical: $3x + 1 = 16$

Step 2: Solve for $x$ : $3x = 15$ $x = 5$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 1: Square both sides to eliminate the radical: $3x + 1 = 16$

Step 2: Solve for $x$ : $3x = 15$ $x = 5$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 1: Square both sides to eliminate the radical: $3x + 1 = 16$

Step 2: Solve for $x$ : $3x = 15$ $x = 5$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 1: Set equal: $x^2 - 3 = 2x - 1$ $x^2 - 2x - 2 = 0$

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$ , the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

$\Delta > 0$ → 2 intersections
$\Delta = 0$ → 1 intersection (tangent)
$\Delta < 0$ → 0 intersections

Step 1: Set equal: $x^2 - 3 = 2x - 1$ $x^2 - 2x - 2 = 0$

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$ , the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

$\Delta > 0$ → 2 intersections
$\Delta = 0$ → 1 intersection (tangent)
$\Delta < 0$ → 0 intersections

Step 1: Set equal: $x^2 - 3 = 2x - 1$ $x^2 - 2x - 2 = 0$

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$ , the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

$\Delta > 0$ → 2 intersections
$\Delta = 0$ → 1 intersection (tangent)
$\Delta < 0$ → 0 intersections

Step 1: Square both sides: $x + 7 = (x+1)^2 = x^2 + 2x + 1$

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$ : $\sqrt{-3 + 7} = \sqrt{4} = 2$ and . Is ? ✗ Extraneous!

$x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and . Is ? ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 1: Square both sides: $x + 7 = (x+1)^2 = x^2 + 2x + 1$

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$ : $\sqrt{-3 + 7} = \sqrt{4} = 2$ and . Is ? ✗ Extraneous!

$x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and . Is ? ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 1: Square both sides: $x + 7 = (x+1)^2 = x^2 + 2x + 1$

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$ : $\sqrt{-3 + 7} = \sqrt{4} = 2$ and . Is ? ✗ Extraneous!

$x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and . Is ? ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 1: Set the equations equal: $x^2 + 2 = 3x + k$ $x^2 - 3x + (2 - k) = 0$

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$

Step 3: Verify: With $k = -\frac{1}{4}$ : $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$

Step 1: Set the equations equal: $x^2 + 2 = 3x + k$ $x^2 - 3x + (2 - k) = 0$

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$

Step 3: Verify: With $k = -\frac{1}{4}$ : $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$

Step 1: Set the equations equal: $x^2 + 2 = 3x + k$ $x^2 - 3x + (2 - k) = 0$

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$

Step 3: Verify: With $k = -\frac{1}{4}$ : $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$

Nonlinear Equations and Functions

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Nonlinear Equations and Functions on the SAT

Beyond Linear: Types of Nonlinear Functions

Quadratic Functions

Absolute Value Functions

Square Root Functions

Rational Functions

Solving Systems with Nonlinear Equations

Linear-Quadratic System

Radical Equations

Solving Equations with Square Roots

Example with Extraneous Solution

Absolute Value Equations

Function Composition and Evaluation

SAT Question Types

Type 1: Solve a Radical Equation

Type 2: Linear-Quadratic System

Type 3: Number of Intersections

Type 4: Absolute Value Equations

Common SAT Mistakes

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4medium

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5Problem 5medium

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6Problem 6medium

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7Problem 7medium

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8Problem 8medium

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9Problem 9medium

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10Problem 10hard

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11Problem 11hard

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12Problem 12hard

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13Problem 13expert

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14Problem 14expert

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15Problem 15expert

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