PSAT 8/9 Math: Nonlinear Equations and Systems
Nonlinear equations introduce new moves beyond basic algebra, like undoing a square root or handling two possible solutions from an absolute value. This free, complete lesson covers square root and radical equations, absolute value equations, squared-binomial equations, and literal equations and nonlinear systems. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the linked quiz below.
1. Square Root and Radical Equations
To solve an equation with a square root, isolate the radical first, then square both sides to undo it. Always check your solution in the original equation, since squaring can sometimes introduce extra, invalid answers.
Worked Example: Solve √(x − 5) = 9 for x.
Square both sides: x − 5 = 81.
Add 5: x = 86
1. Solve √(x − 5) = 9 for x. (see worked example above)
Explanation: Square both sides: x − 5 = 81. Add 5: x = 86.
2. Which value of x satisfies x² = 49?
Explanation: Taking the square root of both sides: x = ±7. Of the choices given, −7 is a valid solution since (−7)² = 49.
3. Solve 72/(x − 3) = x for the positive solution.
Explanation: Multiply both sides by (x − 3): 72 = x(x − 3) → 72 = x² − 3x → x² − 3x − 72 = 0. Factoring: (x − 12)(x + 9) = 0, so x = 12 or x = −9. The positive solution is x = 12.
2. Absolute Value Equations
An absolute value equation like |expression| = k (where k > 0) has two possible cases: the expression equals k, or the expression equals −k. Solve both cases separately to find all solutions.
Worked Example: Solve |x − 20| = 15 for the positive solution.
Case 1: x − 20 = 15 → x = 35.
Case 2: x − 20 = −15 → x = 5.
Both 35 and 5 are positive; the larger solution is x = 35
1. Solve |x − 20| = 15. What are both solutions?
Explanation: Case 1: x − 20 = 15 → x = 35. Case 2: x − 20 = −15 → x = 5. Both are valid solutions.
2. Solve |2x + 6| = 10 for the negative solution.
Explanation: Case 1: 2x + 6 = 10 → x = 2. Case 2: 2x + 6 = −10 → 2x = −16 → x = −8. The negative solution is x = −8.
3. Solve |x| + 5 = 12.
Explanation: Subtract 5 from both sides: |x| = 7. This means x = 7 or x = −7.
3. Squared-Binomial Equations
An equation like (x − a)² = k is solved by taking the square root of both sides, remembering both the positive and negative root, then solving the resulting linear equations.
Worked Example: Given (x − 4)² = 9, find a possible value of x.
Take the square root of both sides: x − 4 = ±3.
Case 1: x − 4 = 3 → x = 7. Case 2: x − 4 = −3 → x = 1.
x = 7 or x = 1
1. Given (x − 4)² = 9, find both possible values of x. (see worked example above)
Explanation: Taking the square root of both sides: x − 4 = ±3, giving x = 7 or x = 1.
2. Given x + 2 = 6 and (x + 2)² = y, find the ordered pair (x, y).
Explanation: From x + 2 = 6, x = 4. Then y = (x + 2)² = 6² = 36. The ordered pair is (4, 36).
3. Given x = 4 and y = (10 − x)², find the product xy.
Explanation: y = (10 − 4)² = 6² = 36. Then xy = 4 × 36 = 144.
4. Literal Equations and Nonlinear Systems
A literal equation has multiple variables, and solving for one means isolating it using the same steps as usual. A nonlinear system (a line and a curve) can have one, two, or no intersection points, each one is a solution shared by both equations.
Worked Example: Given R = PQ, solve for P in terms of R and Q.
Divide both sides by Q: P = R/Q
1. Given R = PQ, solve for P in terms of R and Q. (see worked example above)
Explanation: Divide both sides of R = PQ by Q: P = R/Q.
2. The area of a rectangle is A = lw. Solve for w in terms of A and l.
Explanation: Divide both sides of A = lw by l: w = A/l.
3. A line y = x + 2 and a parabola y = x² intersect at two points. One intersection point is (−1, 1). Verify this point works, then find the other intersection point, given it also satisfies x² − x − 2 = 0.
Explanation: Factoring x² − x − 2 = 0 gives (x − 2)(x + 1) = 0, so x = 2 or x = −1. The other intersection has x = 2, and y = x + 2 = 4, giving (2, 4).
Common Mistakes to Avoid
- Forgetting the negative case when solving an absolute value or squared-binomial equation, both the positive and negative possibilities must be checked.
- Not isolating the radical before squaring in a square root equation, square both sides only after the radical stands alone on one side.
- Forgetting to verify solutions in the original equation after squaring, since squaring can sometimes introduce an "extra" solution that doesn't actually work.
- Dividing by the wrong variable when solving a literal equation, make sure you're isolating the exact variable the question asks for.
- Assuming a line and a curve always intersect at exactly one point. A linear-nonlinear system can have zero, one, or two solutions depending on the shapes involved.
Frequently Asked Questions
Why do I need to check my answer after squaring both sides of an equation?
Squaring both sides of an equation can sometimes turn a false statement into a true one, creating what's called an "extraneous solution." Plugging your answer back into the original equation (before squaring) confirms whether it's actually valid.
How many solutions can an absolute value equation have?
An equation like |expression| = k has two solutions when k is positive, one solution when k is 0, and no solutions when k is negative (since absolute value can never produce a negative result).
Where can I practice more problems like these?
The Nonlinear Equations and Systems quizzes in the PSAT 8/9 Math Question Bank include additional original problems on this topic, along with quizzes covering every other Heart of Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry skill tested on the PSAT 8/9. You can also browse the full PSAT 8/9 Math Question Bank.