11.2 Nonlinear Equations (No Rational or Radical Equations)
Key Takeaways
- A quadratic set to zero is solved by factoring and the zero product property: x^2 - 9x + 18 = 0 becomes (x - 3)(x - 6) = 0, so x = 3 or x = 6.
- If x^2 equals a perfect square, both square roots are solutions: x^2 = 49 gives x = 7 and x = -7. That is not isolating a radical that contains x.
- Integer exponential equations use the same base: 2^x = 32 means x = 5 because 2^5 = 32. Absolute value |x - 3| = 5 means x = 8 or x = -2.
- An elementary nonlinear system such as y = x^2 and y = 4 has two real solutions, the points (2, 4) and (-2, 4).
- Solving rational equations (x in a denominator) and radical equations (isolating a root) is not PSAT 8/9 content; those skills appear on later SAT Suite tests.
A nonlinear equation is not a line: the unknown appears to a power other than 1, in an exponent, or inside an absolute value. College Board's PSAT 8/9 Advanced Math list includes nonlinear equations in one variable and systems of equations in two variables. The same Assessment Framework is explicit that rational and radical equations are not represented on PSAT 8/9. Do not spend test-prep time isolating a square root or clearing an x that sits in a denominator. Those skills appear on later SAT Suite tests (PSAT 10, PSAT/NMSQT, and SAT); see 4.3 How PSAT 8/9 Differs from PSAT 10, PSAT/NMSQT, and the SAT. This section stays on quadratics, integer-level exponentials, absolute value, and elementary mixed systems.
Every Math question allows a calculator. Desmos can graph y = x^2 - 5x + 6 and show x-intercepts 2 and 3. You still need the algebra: factoring, the zero product property, and a substitution check. OpenExamPrep's examples use original numbers; they are not College Board items. Equivalent-form fluency from 11.1 Equivalent Expressions is the engine for factoring.
Quadratic equations: factor and zero product
A quadratic equation can be written ax^2 + bx + c = 0 with a ≠ 0. If it factors as (x - r)(x - s) = 0, the zero product property says a product is zero only when a factor is zero, so x = r or x = s.
Solve x^2 - 9x + 18 = 0.
Numbers that multiply to 18 and add to 9: 3 and 6. Then (x - 3)(x - 6) = 0.
x - 3 = 0 or x - 6 = 0, so x = 3 or x = 6.
Check: 9 - 27 + 18 = 0. 36 - 54 + 18 = 0.
Solve x^2 - 5x + 6 = 0.
(x - 2)(x - 3) = 0, so x = 2 or x = 3.
Check: 4 - 10 + 6 = 0. 9 - 15 + 6 = 0.
Solve x^2 + 7x + 12 = 0.
(x + 3)(x + 4) = 0, so x = -3 or x = -4.
Check: 9 - 21 + 12 = 0. 16 - 28 + 12 = 0.
Solve x^2 - x - 12 = 0.
(x - 4)(x + 3) = 0, so x = 4 or x = -3.
Check: 16 - 4 - 12 = 0. 9 + 3 - 12 = 0.
If the equation is not already equal to zero, move terms first.
Solve x^2 + 3x = 10.
x^2 + 3x - 10 = 0. (x + 5)(x - 2) = 0, so x = -5 or x = 2.
Check: (-5)^2 + 3(-5) = 25 - 15 = 10. 4 + 6 = 10.
GCF first when you can: 2x^2 + 10x + 12 = 0. Factor 2: 2(x^2 + 5x + 6) = 0, so x^2 + 5x + 6 = 0, then (x + 2)(x + 3) = 0, so x = -2 or x = -3.
Check in the original: 2(4) + 10(-2) + 12 = 8 - 20 + 12 = 0. 2(9) + 10(-3) + 12 = 18 - 30 + 12 = 0.
PSAT 8/9 items at this level usually factor with integers or collapse to a perfect square. If a quadratic does not factor nicely, graph both sides in Desmos and read the x-coordinates of the intersections, then still substitute those x-values into the original equation. Completing the square and the quadratic formula are later Algebra 1 tools; they are not required to start this testing point at grades 8 and 9.
Quadratic equations: square roots of perfect squares
If the equation is x^2 = k and k is a perfect square, take square roots: x = √k or x = -√k. Both signs matter. This is not solving a radical equation. The unknown is not trapped under a root that you must isolate; you are reading both square roots of a number.
x^2 = 49. 7^2 = 49 and (-7)^2 = 49, so x = 7 or x = -7.
x^2 = 4. x = 2 or x = -2.
(x - 1)^2 = 16. Then x - 1 = 4 or x - 1 = -4, so x = 5 or x = -3.
Check: (5 - 1)^2 = 16. (-3 - 1)^2 = (-4)^2 = 16.
x^2 - 36 = 0 is the same as x^2 = 36, so x = 6 or x = -6, which also factors as (x - 6)(x + 6) = 0.
If x^2 = 0, the only solution is x = 0. If x^2 = -9, there is no real solution (no real number squares to a negative). PSAT 8/9 stays in real numbers.
Do not turn √(x + 5) = 3 into a PSAT 8/9 drill. Isolating a radical and squaring both sides is later SAT Suite work.
Exponential equations at integer level
Rewrite both sides with the same base, then set the exponents equal. Know the small powers.
| Power | Value |
|---|---|
| 2^0 | 1 |
| 2^1 | 2 |
| 2^2 | 4 |
| 2^3 | 8 |
| 2^4 | 16 |
| 2^5 | 32 |
| 2^6 | 64 |
| 3^3 | 27 |
| 3^4 | 81 |
| 4^3 | 64 |
| 5^3 | 125 |
2^x = 8. 8 is 2^3, so x = 3. Check: 2^3 = 8.
2^x = 32. x = 5, because 2^5 = 32. 2^4 = 16 is too small; 2^6 = 64 is too big.
3^x = 81. 3^4 = 81, so x = 4. Check: 3 · 3 · 3 · 3 = 9 · 9 = 81.
5^x = 1. Any nonzero number to the power 0 is 1, so x = 0. Check: 5^0 = 1.
2^{x+1} = 16. 16 = 2^4, so x + 1 = 4, x = 3. Check: 2^4 = 16.
4^x = 64. Rewrite 4 as 2^2: (2^2)^x = 2^{2x} = 2^6, because 64 = 2^6. Then 2x = 6, x = 3. Check: 4^3 = 64.
These are integer (or simple same-base) items. You do not need logarithms on PSAT 8/9. An equation such as 2^x = 10, which has no integer x, is not the 8/9 version of this skill.
Absolute value equations
|A| = k means the distance from A to 0 is k. If k > 0, two cases: A = k or A = -k. If k = 0, one solution: A = 0. If k < 0, no real solution, because a distance is never negative.
|x - 3| = 5.
x - 3 = 5 or x - 3 = -5, so x = 8 or x = -2.
Check: |8 - 3| = 5. |-2 - 3| = |-5| = 5.
|x| = 7. x = 7 or x = -7.
|2x + 1| = 7.
2x + 1 = 7 or 2x + 1 = -7. 2x = 6 or 2x = -8. x = 3 or x = -4.
Check: |6 + 1| = 7. |-8 + 1| = |-7| = 7.
|x + 4| = 0. Only x = -4. Check: |-4 + 4| = 0.
|x - 1| = -3. No real solution.
A trap is solving only the positive case and keeping x = 8 for |x - 3| = 5. Both numbers sit 5 units from 3 on the number line: 3 + 5 = 8 and 3 - 5 = -2.
Simple systems that include a nonlinear equation
A system can mix a curve and a line, or a curve and a horizontal line. Substitute. Keep the algebra elementary: after substituting you should factor or take square roots of a perfect square.
y = x^2 and y = 4.
x^2 = 4, so x = 2 or x = -2. The solutions are the points (2, 4) and (-2, 4).
Check: 2^2 = 4 and (-2)^2 = 4. Two intersections: a parabola and a horizontal line y = 4.
y = x^2 and y = 2x + 3.
x^2 = 2x + 3 → x^2 - 2x - 3 = 0 → (x - 3)(x + 1) = 0 → x = 3 or x = -1.
When x = 3, y = 2(3) + 3 = 9 (and 3^2 = 9). When x = -1, y = 2(-1) + 3 = 1 (and (-1)^2 = 1).
Points (3, 9) and (-1, 1).
y = x^2 - 4 and y = 5.
x^2 - 4 = 5 → x^2 = 9 → x = 3 or x = -3. Points (3, 5) and (-3, 5). Check: 9 - 4 = 5.
If y = x^2 and y = -1, then x^2 = -1, no real solution. The horizontal line misses the parabola.
Two-linear systems (two lines) live in 10.1 Systems of Two Linear Equations. Here at least one equation is nonlinear.
What this test does not ask you to solve
| In scope on PSAT 8/9 | Out of scope on PSAT 8/9 |
|---|---|
| x^2 - 5x + 6 = 0 | (x + 1)/x = 4 |
| x^2 = 49 | √(x + 5) = 3 |
| 2^x = 8 | 2^x = 10 (needs a logarithm) |
| x - 3 | |
| y = x^2 and y = 4 | A rational system such as 1/x + 1/y = 1 |
You may still evaluate a square root of a perfect square (√49 = 7) or use the Bluebook reference sheet later for geometry. Evaluating is not the same as solving √(expression with x) = number by isolating and squaring.
On SPR, read whether the stem wants both solutions, the positive solution only, or a product. For x = 3 and x = 6, the product is 18. For x = 7 and x = -7, the sum is 0. Type the number the box wants. SPR limits live in 15.1 Student-Produced Response Questions.
Desmos check: graph each side, read intersections. For |x - 3| = 5, graph y = |x - 3| and y = 5; they meet at x = 8 and x = -2. Graphing habits return in 14.3 Reference Sheet, Desmos, and What This Test Does Not Include.
What are the solutions of x^2 - 9x + 18 = 0?
What are the solutions of |x - 3| = 5?
If 2^x = 32, what is x?