Nonlinear Equations and Functions

Solve quadratic, absolute value, and exponential equations.

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Nonlinear Equations and Functions on the SAT

Beyond Linear: Types of Nonlinear Functions

Quadratic Functions

$f(x) = ax^2 + bx + c$ Graph: parabola. Covered in depth in the Quadratic Equations topic.

Absolute Value Functions

$f(x) = |ax + b| + c$ Graph: V-shape. The vertex is at $(-b/a, c)$ .

Square Root Functions

$f(x) = \sqrt{ax + b}$ Graph: Half-parabola (starts at a point, curves right). Domain: $ax + b \geq 0$

Rational Functions

$f(x) = \frac{a}{x - h} + k$ Graph: Hyperbola with asymptotes at $x = h$ and $y = k$ .

Solving Systems with Nonlinear Equations

Systems involving one linear and one nonlinear equation:

Linear-Quadratic System

$y = x + 2 \quad \text{and} \quad y = x^2$

Substitute: $x^2 = x + 2$ $x^2 - x - 2 = 0$ $(x-2)(x+1) = 0$ $x = 2 \text{ or } x = -1$

Solutions: $(2, 4)$ and $(-1, 1)$

Number of solutions:

The line can intersect the parabola at 0, 1, or 2 points
0 points: no solution
1 point: tangent line
2 points: two solutions

Radical Equations

Solving Equations with Square Roots

Strategy: Isolate the radical, then square both sides.

$\sqrt{2x + 3} = 5$ $2x + 3 = 25$ $2x = 22$ $x = 11$

Always check! Squaring can introduce extraneous solutions.

Example with Extraneous Solution

$\sqrt{x + 5} = x - 1$ $x + 5 = (x-1)^2 = x^2 - 2x + 1$ $0 = x^2 - 3x - 4 = (x-4)(x+1)$ $x = 4 \text{ or } x = -1$

Check $x = 4$ : $\sqrt{9} = 3$ and $4 - 1 = 3$ ✓ Check $x = -1$ : $\sqrt{4} = 2$ and $-1 - 1 = -2$ ✗ (extraneous!)

Absolute Value Equations

$|ax + b| = c$

If $c \geq 0$ : Two equations → $ax + b = c$ or $ax + b = -c$ If $c < 0$ : No solution (absolute value can't be negative)

Example: $|2x - 3| = 7$ $2x - 3 = 7 \implies x = 5$ $2x - 3 = -7 \implies x = -2$

Function Composition and Evaluation

For complex function problems:

Read carefully — what specific value or expression are they asking for?
Substitute step by step
Simplify completely

SAT Question Types

Type 1: Solve a Radical Equation

Isolate the radical, square both sides, check for extraneous solutions.

Type 2: Linear-Quadratic System

Substitute and solve the resulting quadratic.

Type 3: Number of Intersections

Use the discriminant of the resulting quadratic to determine 0, 1, or 2 intersections.

Type 4: Absolute Value Equations

Split into two cases.

Common SAT Mistakes

Not checking for extraneous solutions in radical equations
Forgetting there are two cases for absolute value
Errors when squaring both sides — expand $(x-1)^2$ carefully!
Assuming a nonlinear system always has 2 solutions — it could have 0 or 1
Domain errors — $\sqrt{x}$ requires $x \geq 0$

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: $|x - 4| = 6$

💡 Show Solution

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 $x = -2$

2Problem 2easy

❓ Question:

Solve: $|x - 4| = 6$

💡 Show Solution

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 $x = -2$

3Problem 3medium

❓ Question:

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

💡 Show Solution

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: $x = 5$

4Problem 4medium

❓ Question:

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

💡 Show Solution

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: $x = 5$

5Problem 5medium

❓ Question:

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

💡 Show Solution

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

6Problem 6medium

❓ Question:

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

💡 Show Solution

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

7Problem 7hard

❓ Question:

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

💡 Show Solution

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 $-3 + 1 = -2$ . Is $2 = -2$ ? No! ✗ Extraneous!

$x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 + 1 = 3$ . Is $3 = 3$ ? Yes! ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

8Problem 8hard

❓ Question:

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

💡 Show Solution

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 $-3 + 1 = -2$ . Is $2 = -2$ ? No! ✗ Extraneous!

$x = 2$ : $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 + 1 = 3$ . Is $3 = 3$ ? Yes! ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

9Problem 9expert

❓ Question:

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

💡 Show Solution

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: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$ $9 - 8 + 4k = 0$ $1 + 4k = 0$ $k = -\frac{1}{4}$

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

10Problem 10expert

❓ Question:

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

💡 Show Solution

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: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$ $9 - 8 + 4k = 0$ $1 + 4k = 0$ $k = -\frac{1}{4}$

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

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