Nonlinear Equations in One Variable: Practice Questions & Study Guide
Solving quadratic, radical, and rational equations, and understanding the conditions under which extraneous solutions arise.
Understanding Nonlinear Equations in One Variable
Nonlinear equations in one variable include quadratic equations (ax^2 + bx + c = 0), radical equations (sqrt(2x + 1) = x - 1), and rational equations (1/x + 2 = 3/x). Unlike linear equations, these can have two solutions, one solution, or no real solutions—and they require different solving strategies depending on their form. The test tests all three types, with quadratic equations appearing most frequently.
For quadratic equations, the three main solving approaches are factoring (fastest when it works), completing the square (reliable but slower), and the quadratic formula (always works but prone to arithmetic error). The choice depends on the equation: if the coefficient of x^2 is 1 and the constant and linear terms suggest integer factors, try factoring first. If the equation is in vertex form or near it, completing the square is natural. If coefficients are large or irrational, the quadratic formula is the safest bet.
Extraneous solutions are introduced when you raise both sides of an equation to a power or multiply both sides by a variable expression. For radical equations, always square both sides to eliminate the radical, then check each solution in the original equation. If a candidate solution makes the original equation undefined or yields a different value than claimed, it is extraneous and must be discarded. The test tests this by including extraneous solutions among the answer choices—students who don't check lose points here.
Rational equations on the test typically require multiplying through by the LCD to clear denominators, converting to a polynomial equation. Again, solutions that make any original denominator zero are extraneous. Identifying the domain restrictions at the start (noting which values of x make denominators zero) helps you flag potential extraneous solutions before they fool you.
Key Rules & Formulas
Memorize these rules — they come up directly in practice questions.
Zero product property: if (x - a)(x - b) = 0, then x = a or x = b.
(x - 3)(x + 5) = 0 → x = 3 or x = -5.
Quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a) for ax^2 + bx + c = 0.
For x^2 - 5x + 6 = 0: x = (5 ± sqrt(25-24))/2 = (5 ± 1)/2 → x = 3 or x = 2.
Discriminant b^2 - 4ac: positive → 2 solutions; zero → 1 solution; negative → no real solutions.
For x^2 + 4x + 4 = 0: discriminant = 16 - 16 = 0 → one solution x = -2.
To solve a radical equation, isolate the radical then square both sides, then check for extraneous solutions.
sqrt(x + 3) = x - 1: square both sides: x + 3 = x^2 - 2x + 1 → x^2 - 3x - 2 = 0 → check solutions.
For rational equations, multiply both sides by the LCD then solve, discarding solutions that make any denominator zero.
3/(x-2) = 6: multiply by (x-2): 3 = 6(x-2) = 6x-12 → 6x = 15 → x = 5/2. Check: x ≠ 2 ✓.
Nonlinear Equations in One Variable Practice Questions
Select an answer and click Check Answer to reveal the full explanation. Questions go from easiest to hardest.
What are the solutions to x^2 - 9 = 0?
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Correct answer: C. x = 3 or x = -3
Explanation
Add 9: x^2 = 9. Take the square root of both sides: x = ±3. Both 3 and -3 are solutions.
What are the solutions to x^2 + 5x + 6 = 0?
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Correct answer: B. x = -2 or x = -3
Explanation
Factor: (x + 2)(x + 3) = 0. So x = -2 or x = -3. Check: (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 ✓.
If x^2 - 6x = 0, what are the values of x?
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Correct answer: C. x = 0 or x = 6
Explanation
Factor: x(x - 6) = 0. By zero product property: x = 0 or x = 6.
Using the quadratic formula, what are the solutions to 2x^2 - 7x + 3 = 0?
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Correct answer: A. x = 1/2 or x = 3
Explanation
Using the quadratic formula with a=2, b=-7, c=3: discriminant = 49 - 24 = 25. x = (7 ± 5)/4. So x = 12/4 = 3 or x = 2/4 = 1/2.
For what value of k does x^2 + kx + 25 = 0 have exactly one solution?
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Correct answer: D. k = 10 or k = -10
Explanation
Exactly one solution means the discriminant equals zero: k^2 - 4(1)(25) = 0 → k^2 = 100 → k = ±10. Both k = 10 and k = -10 give exactly one solution.
What is the positive solution to sqrt(3x + 1) = x - 1?
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Correct answer: C. x = 5
Explanation
Square both sides: 3x + 1 = (x-1)^2 = x^2 - 2x + 1 → 0 = x^2 - 5x → 0 = x(x-5) → x = 0 or x = 5. Check x=0: sqrt(1) = 0 - 1 → 1 = -1. False, so x = 0 is extraneous. Check x=5: sqrt(16) = 4 = 5-1 = 4 ✓.
How many real solutions does the equation x^2 + 4x + 7 = 0 have?
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Correct answer: A. Zero
Explanation
Discriminant = b^2 - 4ac = 16 - 28 = -12 < 0. A negative discriminant means no real solutions—the parabola does not intersect the x-axis.
If x^2 - 5x + c = 0 has two solutions whose product is 6, what is the value of c?
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Correct answer: B. c = 6
Explanation
By Vieta's formulas, the product of the roots of x^2 - 5x + c = 0 is c/1 = c. Since the product is 6, c = 6. (The sum of roots = 5, consistent with roots 2 and 3: 2×3=6 ✓, 2+3=5 ✓.)
The equation x^2 + bx + 9 = 0 has two equal real roots. What are the possible values of b?
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Correct answer: C. b = 6 or b = -6
Explanation
Two equal real roots means discriminant = 0: b^2 - 4(1)(9) = 0 → b^2 = 36 → b = ±6.
Solve: 4/(x - 2) + 1 = 6/(x - 2). What is the value of x?
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Correct answer: B. x = 4
Explanation
Let u = x - 2. Then 4/u + 1 = 6/u → multiply by u: 4 + u = 6 → u = 2 → x - 2 = 2 → x = 4. Check: 4/(4-2) + 1 = 2 + 1 = 3 and 6/(4-2) = 3 ✓.
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Common Mistakes to Avoid
These are the most frequent errors students make on Nonlinear Equations in One Variable questions. Knowing them in advance prevents costly point losses.
- !Forgetting to check for extraneous solutions after solving radical or rational equations—plugging back in is mandatory.
- !In the quadratic formula, computing -b incorrectly when b is negative: for x^2 - 5x + 4, b = -5 so -b = 5, not -5.
- !Factoring out (x - a) but forgetting that if x = a causes division by zero in a rational equation, it must be excluded.
- !Setting a quadratic equal to a non-zero constant before factoring: (x-2)(x+3) = 4 does NOT mean x - 2 = 4 or x + 3 = 4; expand first.
- !Making an arithmetic error with the discriminant, particularly when a, b, or c are negative.
Strategy Tips: Nonlinear Equations in One Variable
Before solving any radical or rational equation, note the domain restrictions (values that make radicals negative under the sign or denominators zero) so you can immediately flag potential extraneous solutions.
Always check solutions in the ORIGINAL equation, not an intermediate step—checking in a squared or multiplied-through version can validate extraneous solutions.
When the test gives a quadratic in a specific context and asks for a positive solution, the answer is usually positive; the negative root is there as a trap—check the context constraints.
Use the graphing calculator to graph the equation (set equal to zero, graph y = left side - right side) and read off the x-intercepts to confirm your algebraic solutions.
Other Advanced Math Subtopics
Equivalent Expressions
Rewriting algebraic expressions into equivalent forms through factoring, expanding, and applying algebraic identities.
Nonlinear Functions
Understanding polynomial, radical, and rational functions—evaluating them, identifying their key features, and interpreting them in context.
Quadratic Equations and Parabolas
Solving quadratics, identifying the vertex and intercepts of parabolas, and interpreting these features in applied contexts.
Exponential Functions
Modeling growth and decay with exponential functions, interpreting the base and exponent, and comparing exponential to linear growth.
Master Nonlinear Equations in One Variable
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