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Nonlinear equations in one variable

SAT Math Β· Advanced Math Β· Week 5 of the 12-week plan
7

Nonlinear equations in one variable

College Board skill: Nonlinear equations in one variable and systems of equations in two variables
GoalSolve quadratic, radical, rational and absolute value equations, decide how many solutions a quadratic has, and throw away extraneous solutions.
On the test

Advanced Math is about 35% of SAT Math (13–15 of 44 questions). Nonlinear equations in one variable appear in both modules, in multiple-choice and typed-answer form, often with a question such as β€œwhat is the sum of the solutions?” or β€œfor what value of c is there exactly one solution?”.

Key words
quadratic equation Β· an equation that can be written axΒ² + bx + c = 0 with a β‰  0discriminant Β· the number bΒ² βˆ’ 4ac; its sign tells how many real solutions there areextraneous solution Β· an answer found by algebra that does not work in the original equationabsolute value Β· the distance from 0: |βˆ’5| = 5 and |5| = 5
Explanation

Quadratic equations: three tools

First write the equation as axΒ² + bx + c = 0. Then choose a tool. Factoring is fastest when the numbers are small: xΒ² βˆ’ 2x βˆ’ 3 = (x βˆ’ 3)(x + 1) = 0, so x = 3 or x = βˆ’1 (a product is 0 only if one factor is 0). The quadratic formula always works: x = (βˆ’b Β± √(bΒ² βˆ’ 4ac)) / (2a). Completing the square rewrites xΒ² + bx as (x + b/2)Β² βˆ’ (b/2)Β². The solutions are the x-intercepts of the parabola y = axΒ² + bx + c.

xyβˆ’3βˆ’2βˆ’112345βˆ’5βˆ’4βˆ’3βˆ’2βˆ’11234560y = xΒ² βˆ’ 2x βˆ’ 3x = βˆ’1x = 3

The discriminant counts the solutions

The part under the square root, D = bΒ² βˆ’ 4ac, decides everything. If D > 0 the parabola crosses the x-axis twice: two real solutions. If D = 0 it touches the axis once: exactly one solution. If D < 0 it never reaches the axis: no real solutions. The SAT often asks for the value of a constant that makes D equal to 0.

DiscriminantParabola and x-axisReal solutions
D > 0crosses twicetwo
D = 0touches onceone
D < 0never meetsnone

Radical equations and extraneous solutions

To solve √(expression) = something, first isolate the root, then square both sides, solve, and check every answer in the original equation. Squaring can create answers that do not work. For example, √x = βˆ’3 has no solution, because a square root is never negative, but squaring gives x = 9, and √9 = 3 β‰  βˆ’3. Always check, especially when the other side contains x.

Rational equations

Multiply every term by the common denominator to remove the fractions, solve the new equation, and then check that no answer makes a denominator zero. If it does, that answer is extraneous. Example: 1/(x βˆ’ 2) = 3 gives 1 = 3(x βˆ’ 2), so x = 7/3. The value x = 2 is never allowed, because it makes the denominator 0.

Absolute value equations

|A| = k with k > 0 means A = k or A = βˆ’k. So |2x βˆ’ 5| = 9 gives 2x βˆ’ 5 = 9 (x = 7) or 2x βˆ’ 5 = βˆ’9 (x = βˆ’2). On the number line, both answers are 4.5 units from 2.5, the value that makes 2x βˆ’ 5 zero. If k < 0 there is no solution, because an absolute value is never negative. If the right side contains x, as in |x βˆ’ 2| = 2x βˆ’ 7, solve both cases and check each answer, because the right side must not be negative.

βˆ’4βˆ’3βˆ’2βˆ’1012345678910βˆ’22.57
Worked examples
Example 1.
What is the positive solution of 2xΒ² = 3x + 14?
  1. Write in standard form: 2xΒ² βˆ’ 3x βˆ’ 14 = 0.
  2. Factor: (2x βˆ’ 7)(x + 2) = 0.
  3. So x = 7/2 or x = βˆ’2.
  4. The positive solution is 7/2 = 3.5.
Example 2.
For what value of c does xΒ² βˆ’ 8x + c = 0 have exactly one real solution?
xy12345678βˆ’22468100y = xΒ² βˆ’ 8x + 16(4, 0)
  1. One solution means the discriminant is 0.
  2. D = (βˆ’8)Β² βˆ’ 4(1)(c) = 64 βˆ’ 4c.
  3. 64 βˆ’ 4c = 0, so c = 16.
  4. Check: xΒ² βˆ’ 8x + 16 = (x βˆ’ 4)Β², whose only root is 4.
Example 3.
What is the solution to √(2x + 9) = x + 3?
  1. Square both sides: 2x + 9 = xΒ² + 6x + 9.
  2. Move everything to one side: 0 = xΒ² + 4x, so x(x + 4) = 0.
  3. Candidates: x = 0 and x = βˆ’4.
  4. Check x = 0: √9 = 3 and 0 + 3 = 3 βœ“.
  5. Check x = βˆ’4: √1 = 1, but βˆ’4 + 3 = βˆ’1, so it fails.
Trap: x = βˆ’4 is extraneous. Squaring both sides lost the information that √ is never negative.
Example 4.
How many solutions does x/(x βˆ’ 3) = 2 + 3/(x βˆ’ 3) have?
  1. 0
  2. 1
  3. 2
  4. Infinitely many
  1. Multiply every term by x βˆ’ 3: x = 2(x βˆ’ 3) + 3.
  2. x = 2x βˆ’ 6 + 3, so x = 3.
  3. But x = 3 makes the denominators zero, so it is not allowed.
  4. The equation has no solution.
Trap: Stopping at x = 3 and answering β€œ1 solution” is the classic extraneous-solution mistake.
Common traps
  • Not moving everything to one side first(x βˆ’ 2)(x + 5) = 18 does not mean x βˆ’ 2 = 18. Expand and write axΒ² + bx + c = 0 before factoring.
  • Forgetting the Β± or the second solutionxΒ² = 49 gives x = 7 or x = βˆ’7. A quadratic usually has two solutions.
  • Not checking extraneous solutionsAfter squaring or clearing fractions, put every candidate into the original equation.
  • Sign mistakes in the discriminantUse bΒ² βˆ’ 4ac with the signs of b, a and c: for xΒ² βˆ’ 8x + c, bΒ² = (βˆ’8)Β² = 64, not βˆ’64.
  • Absolute value with only one case|A| = k gives two equations: A = k and A = βˆ’k. Check both.
  • Typing a fraction answer wronglyFor a solution such as 7/2, type 7/2 or 3.5. Never 3 1/2. Only the answer is typed, not β€œx =”.
The Desmos way

Graph each side of the equation as a separate line: y = left side and y = right side. The x-coordinates of the crossing points are the solutions. Click each point to read it. For a quadratic set to 0, graph y = the quadratic and click the x-intercepts. A radical or rational equation graphed this way shows extraneous solutions automatically: if the curves do not cross there, the value is not a solution.

  1. Type y = 2x^2
  2. Type y = 3x + 14
  3. Click the two crossing points: x = βˆ’2 and x = 3.5
  4. Type y = √(2x + 9) and y = x + 3 to see that the curves meet only at x = 0

Use Desmos for radical, rational and absolute value equations, and whenever the quadratic does not factor nicely. For simple factorable quadratics, factoring is just as fast.

Open Desmos β†—
Quick check
1
Solve xΒ² βˆ’ 4x βˆ’ 5 = 0.
2
How many real solutions does xΒ² + 2x + 5 = 0 have?
3
Solve √(x βˆ’ 1) = 3.
4
Solve |x + 2| = 7.
Practice set: 10 SAT-style questionsEasy β†’ hard, with typed answers like the real test. Your score is saved in your cabinet.
Start practice β†’

More official practice

Bluebook practice tests β†—Official full-length adaptive tests from the College Board, free SAT question bank β†—Official questions you can filter by skill and difficulty Khan Academy β†—Free official SAT lessons and practice Desmos calculator β†—The same graphing calculator that is built into the test