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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)=ax2+bx+cf(x) = ax^2 + bx + c Graph: parabola. Covered in depth in the Quadratic Equations topic.

Absolute Value Functions

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

Square Root Functions

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

Rational Functions

f(x)=ax−h+kf(x) = \frac{a}{x - h} + k Graph: Hyperbola with asymptotes at x=hx = h and y=ky = k.

Solving Systems with Nonlinear Equations

Systems involving one linear and one nonlinear equation:

Linear-Quadratic System

y=x+2andy=x2y = x + 2 \quad \text{and} \quad y = x^2

Substitute: x2=x+2x^2 = x + 2 x2−x−2=0x^2 - x - 2 = 0 (x−2)(x+1)=0(x-2)(x+1) = 0 x=2 or x=−1x = 2 \text{ or } x = -1

Solutions: (2,4)(2, 4) and (−1,1)(-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.

2x+3=5\sqrt{2x + 3} = 5 2x+3=252x + 3 = 25 2x=222x = 22 x=11x = 11

Always check! Squaring can introduce extraneous solutions.

Example with Extraneous Solution

x+5=x−1\sqrt{x + 5} = x - 1 x+5=(x−1)2=x2−2x+1x + 5 = (x-1)^2 = x^2 - 2x + 1 0=x2−3x−4=(x−4)(x+1)0 = x^2 - 3x - 4 = (x-4)(x+1) x=4 or x=−1x = 4 \text{ or } x = -1

Check x=4x = 4: 9=3\sqrt{9} = 3 and 4−1=34 - 1 = 3 ✓ Check x=−1x = -1: 4=2\sqrt{4} = 2 and −1−1=−2-1 - 1 = -2 ✗ (extraneous!)

Absolute Value Equations

∣ax+b∣=c|ax + b| = c

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

Example: ∣2x−3∣=7|2x - 3| = 7 2x−3=7  ⟹  x=52x - 3 = 7 \implies x = 5 2x−3=−7  ⟹  x=−22x - 3 = -7 \implies x = -2

Function Composition and Evaluation

For complex function problems:

  1. Read carefully — what specific value or expression are they asking for?
  2. Substitute step by step
  3. 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

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: ∣x−4∣=6|x - 4| = 6

💡 Show Solution

Two cases:

Case 1: x−4=6  ⟹  x=10x - 4 = 6 \implies x = 10

Case 2: x−4=−6  ⟹  x=−2x - 4 = -6 \implies x = -2

Check: ∣10−4∣=∣6∣=6|10 - 4| = |6| = 6 ✓ and ∣−2−4∣=∣−6∣=6|-2 - 4| = |-6| = 6 ✓

Answer: x=10x = 10 or x=−2x = -2

2Problem 2easy

❓ Question:

Solve: ∣x−4∣=6|x - 4| = 6

💡 Show Solution

Two cases:

Case 1: x−4=6  ⟹  x=10x - 4 = 6 \implies x = 10

Case 2: x−4=−6  ⟹  x=−2x - 4 = -6 \implies x = -2

Check: ∣10−4∣=∣6∣=6|10 - 4| = |6| = 6 ✓ and ∣−2−4∣=∣−6∣=6|-2 - 4| = |-6| = 6 ✓

Answer: x=10x = 10 or x=−2x = -2

3Problem 3easy

❓ Question:

Solve: ∣x−4∣=6|x - 4| = 6

💡 Show Solution

Two cases:

Case 1: x−4=6  ⟹  x=10x - 4 = 6 \implies x = 10

Case 2: x−4=−6  ⟹  x=−2x - 4 = -6 \implies x = -2

Check: ∣10−4∣=∣6∣=6|10 - 4| = |6| = 6 ✓ and ∣−2−4∣=∣−6∣=6|-2 - 4| = |-6| = 6 ✓

Answer: x=10x = 10 or x=−2x = -2

4Problem 4medium

❓ Question:

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

💡 Show Solution

Step 1: Square both sides to eliminate the radical: 3x+1=163x + 1 = 16

Step 2: Solve for xx: 3x=153x = 15 x=5x = 5

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

Answer: x=5x = 5

5Problem 5medium

❓ Question:

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

💡 Show Solution

Step 1: Square both sides to eliminate the radical: 3x+1=163x + 1 = 16

Step 2: Solve for xx: 3x=153x = 15 x=5x = 5

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

Answer: x=5x = 5

6Problem 6medium

❓ Question:

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

💡 Show Solution

Step 1: Square both sides to eliminate the radical: 3x+1=163x + 1 = 16

Step 2: Solve for xx: 3x=153x = 15 x=5x = 5

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

Answer: x=5x = 5

7Problem 7medium

❓ Question:

How many solutions does the system y=x2−3y = x^2 - 3 and y=2x−1y = 2x - 1 have?

💡 Show Solution

Step 1: Set equal: x2−3=2x−1x^2 - 3 = 2x - 1 x2−2x−2=0x^2 - 2x - 2 = 0

Step 2: Use the discriminant: b2−4ac=(−2)2−4(1)(−2)=4+8=12b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12

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

Answer: 2 solutions

Discriminant shortcut:

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

8Problem 8medium

❓ Question:

How many solutions does the system y=x2−3y = x^2 - 3 and y=2x−1y = 2x - 1 have?

💡 Show Solution

Step 1: Set equal: x2−3=2x−1x^2 - 3 = 2x - 1 x2−2x−2=0x^2 - 2x - 2 = 0

Step 2: Use the discriminant: b2−4ac=(−2)2−4(1)(−2)=4+8=12b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12

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

Answer: 2 solutions

Discriminant shortcut:

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

9Problem 9medium

❓ Question:

How many solutions does the system y=x2−3y = x^2 - 3 and y=2x−1y = 2x - 1 have?

💡 Show Solution

Step 1: Set equal: x2−3=2x−1x^2 - 3 = 2x - 1 x2−2x−2=0x^2 - 2x - 2 = 0

Step 2: Use the discriminant: b2−4ac=(−2)2−4(1)(−2)=4+8=12b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12

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

Answer: 2 solutions

Discriminant shortcut:

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

10Problem 10hard

❓ Question:

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

💡 Show Solution

Step 1: Square both sides: x+7=(x+1)2=x2+2x+1x + 7 = (x+1)^2 = x^2 + 2x + 1

Step 2: Rearrange: 0=x2+2x+1−x−7=x2+x−60 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6

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

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

x=−3x = -3: −3+7=4=2\sqrt{-3 + 7} = \sqrt{4} = 2 and −3+1=−2-3 + 1 = -2. Is 2=−22 = -2? No! ✗ Extraneous!

x=2x = 2: 2+7=9=3\sqrt{2 + 7} = \sqrt{9} = 3 and 2+1=32 + 1 = 3. Is 3=33 = 3? Yes! ✓

Answer: x=2x = 2 only

Key lesson: Always check solutions in radical equations!

11Problem 11hard

❓ Question:

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

💡 Show Solution

Step 1: Square both sides: x+7=(x+1)2=x2+2x+1x + 7 = (x+1)^2 = x^2 + 2x + 1

Step 2: Rearrange: 0=x2+2x+1−x−7=x2+x−60 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6

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

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

x=−3x = -3: −3+7=4=2\sqrt{-3 + 7} = \sqrt{4} = 2 and −3+1=−2-3 + 1 = -2. Is 2=−22 = -2? No! ✗ Extraneous!

x=2x = 2: 2+7=9=3\sqrt{2 + 7} = \sqrt{9} = 3 and 2+1=32 + 1 = 3. Is 3=33 = 3? Yes! ✓

Answer: x=2x = 2 only

Key lesson: Always check solutions in radical equations!

12Problem 12hard

❓ Question:

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

💡 Show Solution

Step 1: Square both sides: x+7=(x+1)2=x2+2x+1x + 7 = (x+1)^2 = x^2 + 2x + 1

Step 2: Rearrange: 0=x2+2x+1−x−7=x2+x−60 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6

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

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

x=−3x = -3: −3+7=4=2\sqrt{-3 + 7} = \sqrt{4} = 2 and −3+1=−2-3 + 1 = -2. Is 2=−22 = -2? No! ✗ Extraneous!

x=2x = 2: 2+7=9=3\sqrt{2 + 7} = \sqrt{9} = 3 and 2+1=32 + 1 = 3. Is 3=33 = 3? Yes! ✓

Answer: x=2x = 2 only

Key lesson: Always check solutions in radical equations!

13Problem 13expert

❓ Question:

For what value of kk does the line y=3x+ky = 3x + k intersect the parabola y=x2+2y = x^2 + 2 at exactly one point?

💡 Show Solution

Step 1: Set the equations equal: x2+2=3x+kx^2 + 2 = 3x + k x2−3x+(2−k)=0x^2 - 3x + (2 - k) = 0

Step 2: For exactly one intersection, the discriminant must equal zero: b2−4ac=0b^2 - 4ac = 0 (−3)2−4(1)(2−k)=0(-3)^2 - 4(1)(2 - k) = 0 9−8+4k=09 - 8 + 4k = 0 1+4k=01 + 4k = 0 k=−14k = -\frac{1}{4}

Step 3: Verify: With k=−14k = -\frac{1}{4}: x2−3x+94=0  ⟹  (x−32)2=0  ⟹  x=32x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}

One solution ✓

Answer: k=−14k = -\frac{1}{4}

14Problem 14expert

❓ Question:

For what value of kk does the line y=3x+ky = 3x + k intersect the parabola y=x2+2y = x^2 + 2 at exactly one point?

💡 Show Solution

Step 1: Set the equations equal: x2+2=3x+kx^2 + 2 = 3x + k x2−3x+(2−k)=0x^2 - 3x + (2 - k) = 0

Step 2: For exactly one intersection, the discriminant must equal zero: b2−4ac=0b^2 - 4ac = 0 (−3)2−4(1)(2−k)=0(-3)^2 - 4(1)(2 - k) = 0 9−8+4k=09 - 8 + 4k = 0 1+4k=01 + 4k = 0 k=−14k = -\frac{1}{4}

Step 3: Verify: With k=−14k = -\frac{1}{4}: x2−3x+94=0  ⟹  (x−32)2=0  ⟹  x=32x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}

One solution ✓

Answer: k=−14k = -\frac{1}{4}

15Problem 15expert

❓ Question:

For what value of kk does the line y=3x+ky = 3x + k intersect the parabola y=x2+2y = x^2 + 2 at exactly one point?

💡 Show Solution

Step 1: Set the equations equal: x2+2=3x+kx^2 + 2 = 3x + k x2−3x+(2−k)=0x^2 - 3x + (2 - k) = 0

Step 2: For exactly one intersection, the discriminant must equal zero: b2−4ac=0b^2 - 4ac = 0 (−3)2−4(1)(2−k)=0(-3)^2 - 4(1)(2 - k) = 0 9−8+4k=09 - 8 + 4k = 0 1+4k=01 + 4k = 0 k=−14k = -\frac{1}{4}

Step 3: Verify: With k=−14k = -\frac{1}{4}: x2−3x+94=0  ⟹  (x−32)2=0  ⟹  x=32x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}

One solution ✓

Answer: k=−14k = -\frac{1}{4}

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❓ Frequently Asked Questions

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 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.