Educerie

Educerie · SAT · Math

Advanced Math · ADV.2 Nonlinear equations in one variable and systems of equations in two variables

Where it is examined
both modules, and reliably among the last questions of a module. Part of the ≈35% of the section that is Advanced Math.
The question this unit answers
the equation has an x², a square root or a fraction with x underneath. Which method, and how many answers should you expect?
Before you start
equivalent expressions, especially factoring and the quadratic formula. This unit is that skill put to work.

What you must be able to do

You must be able toWhat it looks like on the test
Solve a quadratic three waysFactoring, the formula, completing the square
Use the discriminant…for what value of c does the equation have exactly one solution?
Solve radical and rational equationsAnd check for extraneous solutions
Solve a system of a line and a curveSubstitute, then solve the quadratic
Say how many real solutions a system hasWhere the discriminant returns

1The idea in one paragraph

A nonlinear equation can have two solutions, one, or none, and half the questions in this unit are about that count rather than about the values. Factor when it factors, use the formula when it does not, and reach for the discriminant whenever the question says "exactly one" or "no real solutions". For a system, substitute the line into the curve; the result is a quadratic, and its discriminant tells you whether the line cuts, touches or misses.

The built-in graphing calculator answers most of this unit by inspection. Graph it, count the crossings, read the intersections.

2Solving a quadratic

Factor when the numbers are friendly:

x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or x = 3

The formula always works: x = [−b ± √(b² − 4ac)] / 2a. Write a, b and c down before substituting — a dropped minus sign in b is the most common error on this test's hardest algebra.

Completing the square when the question asks for a vertex or a minimum, or when a is 1 and b is even.

Every quadratic set to zero has at most two solutions, and the answer choices reflect that: if you find only one, check whether you divided by something containing x, which is how a solution goes missing.

3The discriminant, b² − 4ac

ValueSolutionsThe graph
PositiveTwo realCrosses the x-axis twice
ZeroExactly oneTouches the axis once
NegativeNone realNever reaches the axis

For what value of c does x² + 8x + c = 0 have exactly one real solution?

b2 − 4ac = 0exactly one solution
82 − 4(1)(c) = 0
64 = 4c
c = 16

This is the single most efficient thing to know in Advanced Math, because "exactly one solution" questions look like they need solving and take ten seconds when they do not.

4Radical equations and extraneous solutions

√(x + 6) = x
x + 6 = x2square both sides
x2 − x − 6 = 0
(x − 3)(x + 2) = 0
x = 3 or x = −2

Now check both in the original:

√(3 + 6) = 3 ✓
√(−2 + 6) = 2, not −2 ✗invented by the squaring

So x = 3 only.

Squaring can create solutions that were never there. Any time you square both sides, checking is part of the method, not an optional extra.

5Rational equations

Multiply through by the denominator, solve, then discard anything that makes a denominator zero.

x/(x − 2) = 3/(x − 2) + 4
x = 3 + 4(x − 2)multiply through by (x − 2)
x = 4x − 5
−3x = −5
x = 5/3x = 2 is excluded; 5/3 is fine

If your answer equals an excluded value, the equation has no solution — a real answer choice on this test.

6Systems: a line and a curve

y = x2 − 2x − 3
y = x + 1
x + 1 = x2 − 2x − 3substitute the line into the curve
x2 − 3x − 4 = 0
(x − 4)(x + 1) = 0
x = 4 or x = −1the points (4, 5) and (−1, 0)

For "how many solutions does the system have", do the substitution and take the discriminant of the resulting quadratic. Two crossings, one touch, or a miss — the same three cases.


Where points are lost

  • Not checking radical solutions in the original equation.
  • Forgetting the ± and reporting one root.
  • Dropping a sign when substituting into the quadratic formula.
  • Dividing both sides by x, which deletes the solution x = 0.
  • Keeping a solution that makes a denominator zero.
  • Solving a "how many solutions" question instead of using the discriminant.
  • Stopping at x when the question wanted the y-coordinate too.

Work it right

  1. Decide what is being asked: values, or a count?
  2. If a count, go to the discriminant.
  3. If values, try factoring, then the formula.
  4. If you squared, check every answer in the original.
  5. If it is a system, substitute first, then solve the quadratic that appears.

Try it

Q1. What are the solutions to x² − 7x + 10 = 0?

A) x = −2 and x = −5 B) x = 2 and x = 5 C) x = 1 and x = 10 D) x = −1 and x = −10

Q2. For what value of k does x² + 10x + k = 0 have exactly one real solution? (Type your answer.)

Q3. What is the solution to √(2x + 3) = x?

A) x = −1 only B) x = 3 only C) x = −1 and x = 3 D) No real solution

Q4. The system y = x² + 3 and y = 2x + 4 has how many real solutions?

A) Zero B) Exactly one C) Exactly two D) Infinitely many

Q5. If 2x² − 5x − 3 = 0 and x > 0, what is the value of x?

A) 1/2 B) 3 C) −1/2 D) 6

In one breath

Count first, solve second: if the question says "exactly one" or "no real solutions", the discriminant b² − 4ac answers it in ten seconds. When you do solve, factor if it factors and use the formula if it does not, writing a, b and c down before you substitute. Squaring both sides of a radical equation invents solutions, so check every answer in the original — and for a line meeting a curve, substitute and take the discriminant of the quadratic that appears.

Answers

Q1. B. two numbers multiplying to 10 and adding to −7 (x − 2)(x − 5) = 0. A has both signs wrong. C and D use factors of 10 that do not add to −7.

Q2. 25. set the discriminant to zero

b2 − 4ac = 0
102 − 4(1)(k) = 0
k = 25

The graph then touches the x-axis at x = −5 instead of crossing it.

Q3. B — x = 3 only. solve, then test in the original

√(2x + 3) = x
2x + 3 = x2
x2 − 2x − 3 = 0
x = 3 or x = −1
√9 = 3 ✓test in the original
√1 = 1, not −1 ✗

C keeps the extraneous root, which is exactly what this question is built to catch.

Q4. C — exactly two. substitute, then use the discriminant

x2 + 3 = 2x + 4
x2 − 2x − 1 = 0
b2 − 4ac = 4 + 4 = 8positive, so two crossings

B would need a discriminant of zero.

Q5. B — 3. factor, then use the condition

2x2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
x = −1/2 or x = 3
x > 0 → x = 3

C is the rejected root; A is its size with the sign changed.


Educerie · written from the published College Board* Assessment Framework for the Digital SAT Suite *(Math, Advanced Math, skill/knowledge testing point "Nonlinear equations in one variable and systems of equations in two variables"). All questions and explanations are original Educerie text. Last reviewed 12 September 2026.

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