Nonlinear Equations and Systems

Nonlinear Equations in One Variable and Systems of Equations in Two Variables

SAT Math | Advanced Math | Foundation to Advanced

Nonlinear equations are equations that cannot be simplified into a linear equation. They may contain squared variables, higher powers, square roots, variable denominators, or absolute values. On the SAT, you must be able to solve these equations, recognize the number of real solutions, check for invalid answers, and solve systems in which at least one equation is nonlinear.

1. Understanding Nonlinear Equations

A linear equation in one variable has the general form:

ax + b = 0

where a is nonzero.

For example:

3x + 6 = 0

This equation has one solution.

A nonlinear equation involves a relationship that cannot be reduced to this linear form.

Examples include:

x² − 9 = 0

x³ − 8 = 0

√(x + 5) = 4

1/x = 2

|x − 3| = 5

Unlike a linear equation in one variable, a nonlinear equation can have multiple real solutions.

For example:

x² = 16

Taking square roots gives:

x = 4 or x = −4

Both numbers satisfy the original equation.

Key idea: Always check whether a nonlinear equation has more than one possible solution.

On the SAT, nonlinear equations may be presented as equations, word problems, graphs, or parts of a system of equations.

2. Solving Quadratic Equations by Factoring

A quadratic equation is an equation that can be written in the form:

ax² + bx + c = 0

where a ≠ 0.

Quadratic equations contain a squared variable.

For example:

x² − 5x + 6 = 0

One method for solving a quadratic equation is factoring.

The goal is to rewrite the quadratic expression as a product.

The zero-product property

If:

A × B = 0

then at least one factor must equal zero.

Therefore:

A = 0 or B = 0

This principle allows us to solve factored quadratic equations.

Worked Example 1 — Quadratic factoring

Difficulty: Easy

Solve:

x² − 5x + 6 = 0

Step 1: Find two numbers whose product is 6 and whose sum is −5.

The numbers are −2 and −3.

Step 2: Factor the quadratic.

(x − 2)(x − 3) = 0

Step 3: Set each factor equal to zero.

x − 2 = 0

or

x − 3 = 0

Step 4: Solve each equation.

x = 2 or x = 3

Final answers: 2 and 3

Check:

For x = 2:

4 − 10 + 6 = 0

For x = 3:

9 − 15 + 6 = 0

Both are valid.

When the leading coefficient is greater than 1

Consider:

2x² − 7x + 3 = 0

Factor:

(2x − 1)(x − 3) = 0

Solve:

2x − 1 = 0 gives x = 1/2.

x − 3 = 0 gives x = 3.

Final answers: 1/2 and 3

SAT tip: Before using a longer method, check whether the quadratic can be factored easily.

3. Solving Quadratic Equations Using Square Roots

Some quadratic equations are easier to solve using square roots.

Consider:

x² = 25

Taking square roots gives:

x = ±5

The symbol ± means positive or negative.

Both 5 and −5 are solutions because:

5² = 25

(−5)² = 25

Worked Example 2 — Square root method

Difficulty: Easy

Solve:

3(x − 2)² = 48

Step 1: Divide both sides by 3.

(x − 2)² = 16

Step 2: Take both square roots.

x − 2 = ±4

Step 3: Solve each possibility.

x − 2 = 4 gives x = 6.

x − 2 = −4 gives x = −2.

Final answers: 6 and −2

Important SAT rule

If:

x² = a

and a > 0, there are two real solutions.

If a = 0, there is one real solution.

If a < 0, there are no real solutions.

For example:

x² = −9

has no real solution because the square of a real number cannot be negative.

Common mistake: Forgetting the negative solution when taking square roots.

4. The Quadratic Formula

Not every quadratic equation factors easily.

The quadratic formula provides a general method for solving:

ax² + bx + c = 0

The formula is:

x = (−b ± √(b² − 4ac)) / (2a)

where a ≠ 0.

You should be able to identify a, b, and c correctly.

Worked Example 3 — Quadratic formula

Difficulty: Medium

Solve:

2x² − 3x − 2 = 0

Step 1: Identify the coefficients.

a = 2

b = −3

c = −2

Step 2: Substitute into the formula.

x = [−(−3) ± √((−3)² − 4(2)(−2))] / [2(2)]

Step 3: Simplify.

x = [3 ± √(9 + 16)] / 4

x = (3 ± 5) / 4

Step 4: Find both solutions.

First solution:

x = (3 + 5) / 4 = 2

Second solution:

x = (3 − 5) / 4 = −1/2

Final answers: 2 and −1/2

SAT tip

The quadratic formula works even when factoring is difficult.

However, factoring is often quicker when the coefficients are small and the roots are simple.

5. Completing the Square

Completing the square rewrites a quadratic expression into a perfect-square form.

For example:

x² + 6x + 9 = (x + 3)²

A useful identity is:

x² + 2ax + a² = (x + a)²

To complete the square for x² + bx, add:

(b/2)²

Worked Example 4 — Completing the square

Difficulty: Medium

Solve:

x² + 6x − 7 = 0

Step 1: Move the constant.

x² + 6x = 7

Step 2: Find half of the x-coefficient.

6/2 = 3

Step 3: Square the result.

3² = 9

Step 4: Add 9 to both sides.

x² + 6x + 9 = 16

Step 5: Factor the perfect square.

(x + 3)² = 16

Step 6: Take both square roots.

x + 3 = ±4

Step 7: Solve.

x = 1 or x = −7

Final answers: 1 and −7

When completing the square is useful

Use this method when:

• A quadratic does not factor easily.

• You need to recognize a perfect-square structure.

• You need to identify a minimum or maximum.

• You want to rewrite a quadratic into vertex form.

6. Determining the Number of Real Solutions

SAT questions sometimes ask how many real solutions a quadratic equation has without asking you to calculate them.

Use the discriminant.

For:

ax² + bx + c = 0

the discriminant is:

D = b² − 4ac

The value of D determines the number of distinct real solutions.

Worked Example 5 — Number of solutions

Difficulty: Medium

How many real solutions does the equation have?

x² − 4x + 7 = 0

Step 1: Identify the coefficients.

a = 1

b = −4

c = 7

Step 2: Calculate the discriminant.

D = (−4)² − 4(1)(7)

D = 16 − 28

D = −12

Step 3: Interpret the result.

Because D < 0, there are no real solutions.

Final answer: 0 real solutions

Worked Example 6 — Unknown constant

Difficulty: Hard

For what value of k does this quadratic equation have exactly one distinct real solution?

x² − 6x + k = 0

Step 1: Exactly one real solution means D = 0.

Step 2: Identify the coefficients.

a = 1

b = −6

c = k

Step 3: Write the discriminant.

(−6)² − 4(1)(k) = 0

Step 4: Simplify.

36 − 4k = 0

Step 5: Solve.

4k = 36

k = 9

Final answer: 9

SAT tip: If a quadratic has exactly one distinct real solution, its graph touches the x-axis at one point.

7. Solving Polynomial Equations

Polynomial equations may contain higher powers such as x³ or x⁴.

The SAT often tests polynomial equations written in factored form.

Worked Example 7 — Factored polynomial

Difficulty: Medium

Solve:

(x − 2)(x + 3)(2x − 1) = 0

Step 1: Apply the zero-product property.

x − 2 = 0

or

x + 3 = 0

or

2x − 1 = 0

Step 2: Solve each equation.

x = 2

x = −3

x = 1/2

Final answers: −3, 1/2, and 2

Repeated roots

Consider:

(x − 4)²(x + 1) = 0

The solutions are:

x = 4

x = −1

Although the factor x − 4 appears twice, x = 4 is still one distinct solution value.

Important: The number of factors is not always the same as the number of distinct real solutions.

8. Solving Absolute Value Equations

The absolute value of a number is its distance from zero on the number line.

Examples:

|5| = 5

|−5| = 5

Absolute value is never negative.

For an equation:

|A| = b

where b > 0:

A = b or A = −b

Worked Example 8 — Absolute value

Difficulty: Medium

Solve:

|2x − 5| = 7

Step 1: Create two possible equations.

2x − 5 = 7

or

2x − 5 = −7

Step 2: Solve the first equation.

2x = 12

x = 6

Step 3: Solve the second equation.

2x = −2

x = −1

Final answers: 6 and −1

Special cases

If:

|x − 3| = 0

then x = 3.

If:

|x − 3| = −5

there is no solution because an absolute value cannot be negative.

SAT tip: Before splitting an absolute value equation into two cases, check whether the right side is positive, zero, or negative.

9. Solving Radical Equations

A radical equation contains a variable inside a root.

For example:

√(x + 5) = 4

To solve, isolate the radical and then raise both sides to the appropriate power.

Worked Example 9 — Square root equation

Difficulty: Easy

Solve:

√(x + 5) = 4

Step 1: Square both sides.

x + 5 = 16

Step 2: Subtract 5.

x = 11

Step 3: Check in the original equation.

√(11 + 5) = √16 = 4

Final answer: 11

Extraneous solutions

Squaring both sides may create an answer that does not satisfy the original equation.

Such an answer is called an extraneous solution.

Worked Example 10 — Checking solutions

Difficulty: Hard

Solve:

√(x + 5) = x − 1

Step 1: Since a square root is nonnegative, the right side must satisfy x − 1 ≥ 0.

Therefore x ≥ 1.

Step 2: Square both sides.

x + 5 = (x − 1)²

Step 3: Expand.

x + 5 = x² − 2x + 1

Step 4: Move all terms to one side.

x² − 3x − 4 = 0

Step 5: Factor.

(x − 4)(x + 1) = 0

Step 6: Find candidate solutions.

x = 4 or x = −1

Step 7: Check the original equation.

For x = 4:

√9 = 3

4 − 1 = 3

Valid.

For x = −1:

√4 = 2

−1 − 1 = −2

Invalid.

Final answer: 4

The value −1 is extraneous.

Important: Always check candidate solutions in the original radical equation.

10. Solving Rational Equations

A rational equation contains a variable in a denominator.

For example:

(x + 2)/(x − 1) = 3

Before solving, identify values that make any denominator equal zero.

For this equation:

x ≠ 1

Worked Example 11 — Rational equation

Difficulty: Medium

Solve:

(x + 2)/(x − 1) = 3

Step 1: Identify the restriction.

x ≠ 1

Step 2: Multiply both sides by x − 1.

x + 2 = 3(x − 1)

Step 3: Expand.

x + 2 = 3x − 3

Step 4: Rearrange.

5 = 2x

Step 5: Solve.

x = 5/2

Step 6: Check that x is not excluded.

5/2 ≠ 1

Final answer: 5/2

Worked Example 12 — An excluded solution

Difficulty: Hard

Solve:

(x² − 4)/(x − 2) = 4

Step 1: Identify the restriction.

x ≠ 2

Step 2: Factor the numerator.

(x − 2)(x + 2)/(x − 2) = 4

Step 3: Cancel the common factor for x ≠ 2.

x + 2 = 4

Step 4: Solve.

x = 2

Step 5: Check the restriction.

The original denominator becomes zero at x = 2.

Therefore x = 2 is not allowed.

Final answer: No solution

SAT tip: Canceling a factor does not remove the original domain restriction.

11. Systems of Linear and Nonlinear Equations

A system can contain a linear equation and a nonlinear equation.

For example:

y = x + 1

y = x² − 3x + 1

A solution is an ordered pair that satisfies both equations.

The easiest method is often substitution.

Worked Example 13 — Line and parabola

Difficulty: Medium

Solve the system:

y = x + 1

y = x² − 3x + 1

Step 1: Set the two expressions for y equal.

x + 1 = x² − 3x + 1

Step 2: Rearrange.

x² − 4x = 0

Step 3: Factor.

x(x − 4) = 0

Step 4: Solve.

x = 0 or x = 4

Step 5: Find each corresponding y-value.

When x = 0:

y = 0 + 1 = 1

When x = 4:

y = 4 + 1 = 5

Final answers: (0, 1) and (4, 5)

The line and parabola intersect at two points.

Number of intersections

A line and a parabola can have:

• Two distinct real intersection points.

• Exactly one real intersection point.

• No real intersection points.

After substitution, use the resulting quadratic equation to determine which case applies.

12. Systems Involving Circles

A nonlinear system may contain a circle.

A circle centered at the origin has the equation:

x² + y² = r²

where r is the radius.

Worked Example 14 — Line and circle

Difficulty: Medium–Hard

Solve:

x² + y² = 25

y = 3

Step 1: Substitute y = 3 into the circle equation.

x² + 3² = 25

Step 2: Simplify.

x² + 9 = 25

Step 3: Subtract 9.

x² = 16

Step 4: Take both square roots.

x = 4 or x = −4

Step 5: Use y = 3.

Final answers: (4, 3) and (−4, 3)

The horizontal line intersects the circle at two points.

SAT tip

When a system includes a circle and a line, substitution is often the simplest approach.

13. Systems with Exactly One Intersection

A line may touch a parabola at exactly one point.

This occurs when the equation obtained by substitution has one distinct real solution.

Worked Example 15 — Finding an unknown constant

Difficulty: Hard

The system below has exactly one solution:

y = x²

y = 6x − k

What is k?

Step 1: Set the equations equal.

x² = 6x − k

Step 2: Rearrange.

x² − 6x + k = 0

Step 3: Exactly one real solution means the discriminant is zero.

D = b² − 4ac = 0

Step 4: Substitute.

(−6)² − 4(1)(k) = 0

36 − 4k = 0

Step 5: Solve.

k = 9

Final answer: 9

When k = 9:

x² − 6x + 9 = 0

(x − 3)² = 0

The only x-coordinate is 3.

The corresponding y-coordinate is 9.

Therefore, the system has exactly one intersection at (3, 9).

14. Real-World Nonlinear Equations

Nonlinear equations can describe areas, projectile motion, changing dimensions, and other relationships.

Worked Example 16 — Rectangle area

Difficulty: Medium

A rectangle has a width of x meters and a length of x + 5 meters.

Its area is 84 square meters.

What is the width?

Step 1: Use the area formula.

Area = Length × Width

Step 2: Write the equation.

x(x + 5) = 84

Step 3: Expand.

x² + 5x = 84

Step 4: Rearrange.

x² + 5x − 84 = 0

Step 5: Factor.

(x + 12)(x − 7) = 0

Step 6: Solve.

x = −12 or x = 7

Step 7: Interpret the context.

A rectangle cannot have a negative width.

Therefore x = 7.

Final answer: 7 meters

Worked Example 17 — Height of an object

Difficulty: Hard

The height h, in meters, of an object above the ground is modeled by:

h(t) = −5t² + 20t + 1

where t is the time in seconds.

At which times is the object exactly 16 meters above the ground?

Step 1: Set h(t) = 16.

−5t² + 20t + 1 = 16

Step 2: Move all terms to one side.

−5t² + 20t − 15 = 0

Step 3: Divide by −5.

t² − 4t + 3 = 0

Step 4: Factor.

(t − 1)(t − 3) = 0

Step 5: Solve.

t = 1 or t = 3

Final answers: 1 second and 3 seconds

The object reaches the same height once while rising and once while falling.

15. SAT Strategies and Common Mistakes

Strategy 1 — Identify the equation type

Before solving, determine whether the equation is quadratic, polynomial, radical, rational, or absolute value.

Different equation types may require different methods.

Strategy 2 — Factor when possible

Factoring is often the fastest way to solve quadratic and polynomial equations with simple factors.

Strategy 3 — Use the discriminant

If the question asks for the number of real solutions of a quadratic, calculate:

D = b² − 4ac

You may not need to find the roots.

Strategy 4 — Check extraneous solutions

Squaring both sides or multiplying by variable-containing denominators can introduce invalid candidate solutions.

Always check answers in the original equation.

Strategy 5 — Remember both square roots

If x² = 25, the real solutions are x = 5 and x = −5.

Strategy 6 — Solve nonlinear systems by substitution

When y is isolated in both equations, set the expressions equal.

Then solve the resulting equation.

Strategy 7 — Interpret contextual restrictions

Negative lengths, impossible quantities, and other values outside the problem's domain must be rejected.

Strategy 8 — Use Desmos strategically

The Digital SAT includes an embedded Desmos graphing calculator.

Desmos can help you visualize quadratic equations, circle intersections, and systems.

However, algebra is often faster when factoring or substitution is straightforward.

Common Mistake 1 — Losing a solution

Incorrect:

x² = 36 gives only x = 6.

Correct:

x = ±6.

Common Mistake 2 — Using the wrong quadratic coefficients

For:

2x² − 3x − 2 = 0

a = 2, b = −3, and c = −2.

The signs are part of the coefficients.

Common Mistake 3 — Misinterpreting the discriminant

D > 0 means two distinct real solutions.

D = 0 means one distinct real solution.

D < 0 means no real solutions.

Common Mistake 4 — Accepting an extraneous radical solution

Always substitute candidate solutions back into the original radical equation.

Common Mistake 5 — Ignoring undefined denominators

Values making an original denominator zero are never valid solutions.

Common Mistake 6 — Forgetting both absolute value cases

For |A| = b with b > 0, solve A = b and A = −b.

Common Mistake 7 — Solving only one equation in a system

A solution must satisfy both equations.

Common Mistake 8 — Finding x but forgetting y

If the question asks for the ordered pair, determine both coordinates.

Common Mistake 9 — Assuming every nonlinear equation has two real solutions

A quadratic can have zero, one, or two distinct real solutions.

Higher-degree polynomials can have more possibilities.

Common Mistake 10 — Keeping impossible contextual solutions

If an algebraic calculation produces a negative width, reject it when the width must be positive.

16. Check Your Understanding

Question 1 — Quadratic factoring

Solve:

x² − 7x + 12 = 0

A. 2 and 6

B. 3 and 4

C. −3 and −4

D. 1 and 12

Correct answer: B

Factor:

(x − 3)(x − 4) = 0

Therefore x = 3 or x = 4.

Question 2 — Number of real solutions

How many distinct real solutions does the equation have?

x² + 4x + 8 = 0

A. Zero

B. One

C. Two

D. Infinitely many

Correct answer: A

D = 4² − 4(1)(8)

D = 16 − 32 = −16

Since D < 0, there are no real solutions.

Question 3 — Absolute value

Solve:

|x − 2| = 5

A. 7 only

B. −3 only

C. −3 and 7

D. 3 and 7

Correct answer: C

x − 2 = 5 gives x = 7.

x − 2 = −5 gives x = −3.

Question 4 — Radical equation

Solve:

√(x + 4) = 5

Correct answer: 21

Square both sides.

x + 4 = 25

x = 21

Question 5 — Nonlinear system

Solve:

y = x²

y = 2x + 3

Correct answer: (−1, 1) and (3, 9)

Set equations equal:

x² = 2x + 3

x² − 2x − 3 = 0

(x − 3)(x + 1) = 0

Therefore x = 3 or x = −1.

When x = 3, y = 9.

When x = −1, y = 1.

17. Lesson Summary

Nonlinear equations cannot be simplified into linear equations.

Quadratic equations can often be solved by factoring, square roots, completing the square, or the quadratic formula.

The discriminant determines the number of distinct real solutions of a quadratic equation.

Polynomial equations in factored form can be solved using the zero-product property.

Absolute value equations may require solving two separate cases.

Radical equations may produce extraneous solutions after squaring.

Rational equations require checking denominator restrictions.

A nonlinear system can have multiple real intersection points.

Substitution is often the most effective method for solving systems involving a line and a quadratic or circle.

Real-world nonlinear models must be interpreted using appropriate restrictions on the variables.

You are now ready to complete the Nonlinear Equations and Systems practice questions in Baloo Prep.