Nonlinear Equations in One Variable and Systems of Equations

This topic covers solving nonlinear equations and systems of equations, which is one of the most heavily tested areas in the Advanced Math domain of the Digital SAT. You'll need to solve quadratic equations using multiple methods, work with absolute value, rational equations, radical equations, and polynomial equations in factored form, tackle systems that pair linear equations with nonlinear ones, analyze how many solutions a quadratic has, and rearrange multi-variable formulas. Expect roughly 5–8 questions across the two math modules that draw directly on these skills.

Solving Quadratic Equations

Quadratic equations are equations where the highest power of the variable is 2, typically written in standard form $ax^2 + bx + c = 0$. The SAT presents these in many forms, and choosing the right method matters for speed.

Factoring

Factoring is the fastest approach when it works. It relies on the zero product property: if two things multiply to zero, at least one of them equals zero.

When $a = 1$: Find two numbers that multiply to $c$ and add to $b$.

Example: Solve $x^2 - 5x + 6 = 0$

You need two numbers that multiply to 6 and add to $-5$. That's $-2$ and $-3$.

$( x - 2)(x - 3) = 0$

$x = 2 \text{ or } x = 3$

When $a \neq 1$: Find two numbers that multiply to $ac$ and add to $b$, then factor by grouping.

Example: Solve $3x^2 + 10x + 8 = 0$

Multiply $a \cdot c = 3 \cdot 8 = 24$. Find two numbers that multiply to 24 and add to 10: that's 4 and 6.

$3x^2 + 4x + 6x + 8 = 0$ $x(3x + 4) + 2(3x + 4) = 0$ $(3x + 4)(x + 2) = 0$ $x = -\frac{4}{3} \text{ or } x = -2$

The SAT also gives quadratics in non-standard forms. If you see $x^2 = 9x - 14$, rearrange to $x^2 - 9x + 14 = 0$ before factoring. If you see $(2x + 1)^2 = 49$, recognize the algebraic structure and take the square root of both sides directly: $2x + 1 = \pm 7$, giving $x = 3$ or $x = -4$.

The Quadratic Formula

The quadratic formula works on every quadratic equation, including ones that don't factor neatly. For $ax^2 + bx + c = 0$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example: Solve $2x^2 + 3x - 7 = 0$

Identify: $a = 2$, $b = 3$, $c = -7$

$x = \frac{-3 \pm \sqrt{9 - 4(2)(-7)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 56}}{4} = \frac{-3 \pm \sqrt{65}}{4}$

If the SAT asks for the positive solution, pick the "plus" version: $x = \frac{-3 + \sqrt{65}}{4}$.

Watch your signs carefully. The most common error is mishandling negatives when $b$ or $c$ is negative.

Completing the Square

Completing the square converts a quadratic into the form $(x + h)^2 = k$. This is the most efficient method when the problem specifically asks for vertex form or when the coefficient of $x$ is even (making the arithmetic clean).

Example: Solve $x^2 - 8x + 5 = 0$

$x^2 - 8x = -5$

Take half of $-8$, which is $-4$, and square it: $16$. Add to both sides:

$x^2 - 8x + 16 = -5 + 16$

$(x - 4)^2 = 11$

$x - 4 = \pm\sqrt{11}$

$x = 4 \pm \sqrt{11}$

If $a \neq 1$, divide everything by $a$ first. For $2x^2 + 12x + 7 = 0$, divide by 2 to get $x^2 + 6x + \frac{7}{2} = 0$, then proceed.

The Discriminant and Number of Solutions

The discriminant is the expression under the square root in the quadratic formula: $b^2 - 4ac$. It tells you how many real solutions exist without requiring you to solve.

• $b^2 - 4ac > 0$: Two distinct real solutions
• $b^2 - 4ac = 0$: Exactly one real solution (a repeated root)
• $b^2 - 4ac < 0$: No real solutions

The SAT frequently tests this with a parameter. The question gives you a quadratic with an unknown constant and asks what value produces a specific number of solutions.

Example: The equation $x^2 + bx + 9 = 0$ has exactly one real solution. What could be the value of $b$?

Set the discriminant equal to zero:

$b^2 - 4(1)(9) = 0$

$b^2 - 36 = 0$

$b^2 = 36$ $b = 6 \text{ or } b = -6$

Harder Example: For what values of $k$ does $2x^2 - 4x + k = 0$ have no real solutions?

$(-4)^2 - 4(2)(k) < 0$

$16 - 8k < 0$

$16 < 8k$ $k > 2$

Any value of $k$ greater than 2 means no real solutions.

Absolute Value, Rational, and Radical Equations

Absolute Value Equations

An absolute value equation like $|3x - 5| = 7$ means the expression inside can equal 7 or $-7$. Split into two cases:

$3x - 5 = 7 \quad \text{or} \quad 3x - 5 = -7$

$3x = 12 \quad \text{or} \quad 3x = -2$ $x = 4 \quad \text{or} \quad x = -\frac{2}{3}$

If the equation equals a negative number, like $|2x + 1| = -3$, there's no solution. Absolute value can never be negative.

Rational Equations

Rational equations have variables in denominators. Multiply both sides by the denominator to clear fractions, then solve. Always check for extraneous solutions (values that make a denominator zero).

Example: Solve $\frac{x^2 - 4}{x - 2} = 5$

Multiply both sides by $(x - 2)$:

$x^2 - 4 = 5(x - 2)$

$x^2 - 4 = 5x - 10$

$x^2 - 5x + 6 = 0$

$(x - 2)(x - 3) = 0$

$x = 2 \text{ or } x = 3$

But $x = 2$ makes the original denominator zero, so it's extraneous. The only solution is $x = 3$.

Radical Equations

Isolate the radical, then square both sides. Check for extraneous solutions here too, since squaring can introduce false answers.

Example: Solve $\sqrt{2x + 3} = x$

Square both sides:

$2x + 3 = x^2$ $x^2 - 2x - 3 = 0$

$(x - 3)(x + 1) = 0$

$x = 3 \text{ or } x = -1$

Check $x = 3$: $\sqrt{6 + 3} = \sqrt{9} = 3$ ✓ Check $x = -1$: $\sqrt{-2 + 3} = \sqrt{1} = 1 \neq -1$ ✗

Only $x = 3$ works.

Polynomial Equations in Factored Form

When a polynomial equation is already in factored form, apply the zero product property directly. The SAT won't ask you to factor a cubic from scratch, but it will give you something like:

Example: Solve $(x - 4)(x + 1)(2x - 3) = 0$

Set each factor to zero:

$x - 4 = 0 \Rightarrow x = 4$

$x + 1 = 0 \Rightarrow x = -1$ $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$

Three solutions. If the question asks "what is the sum of all solutions," add them: $4 + (-1) + \frac{3}{2} = \frac{9}{2}$.

Sometimes polynomial equations appear with repeated factors like $(x - 2)^2(x + 5) = 0$. This still gives $x = 2$ and $x = -5$, but $x = 2$ is a repeated root. The number of distinct solutions is 2.

Systems of Nonlinear Equations

Systems of nonlinear equations on the SAT typically pair a linear equation with a quadratic. The standard approach: solve the linear equation for one variable, then substitute into the nonlinear equation.

Example: Solve the system: $y = x + 1$ $y = x^2 - 3x + 5$

Substitute the first equation into the second:

$x + 1 = x^2 - 3x + 5$

$0 = x^2 - 4x + 4$

$0 = (x - 2)^2$

$x = 2$

Then $y = 2 + 1 = 3$. The solution is $(2, 3)$.

Because $(x - 2)^2 = 0$ gave one repeated root, this system has exactly one solution. Graphically, the line is tangent to the parabola at that point. If the resulting quadratic had two distinct roots, the line would intersect the parabola at two points. If the discriminant were negative, the line and parabola wouldn't intersect at all. This connection between the algebra and the graphs is something the SAT tests directly.

Solving for a Variable of Interest

Some questions give you a formula with multiple variables and ask you to isolate one. Treat every other variable as a constant and use normal algebra.

Example: Given $A = \pi r^2 h$, solve for $r$.

$\frac{A}{\pi h} = r^2$ $r = \sqrt{\frac{A}{\pi h}}$

(Take the positive root since $r$ represents a physical measurement.)

Trickier Example: Given $y = 3x^2 + 2x$, solve for $x$ in terms of $y$.

Rearrange: $3x^2 + 2x - y = 0$. This is a quadratic in $x$ where $a = 3$, $b = 2$, $c = -y$. Apply the quadratic formula:

$x = \frac{-2 \pm \sqrt{4 + 12y}}{6}$

What to Watch For on Test Day

1. Check for extraneous solutions whenever you solve rational equations or radical equations. Squaring both sides or multiplying by a variable expression can create false answers.

2. Recognize structural shortcuts. If you see $(3x + 1)^2 = 25$, don't expand it. Take the square root directly. If you see $x^4 - 5x^2 + 4 = 0$, treat $x^2$ as a single variable and factor.

3. Use the discriminant strategically. When a question asks how many solutions exist or what value of a parameter gives a certain number of solutions, go straight to $b^2 - 4ac$. Don't waste time solving the full equation.

4. For systems, substitute rather than guess. Solve the linear equation for $y$ (or $x$), plug into the quadratic, and solve the resulting equation. The number of solutions to that equation tells you how many intersection points exist.

5. Sign errors are the top trap. When using the quadratic formula, be especially careful with negative values of $b$ and $c$. Writing out $b^2 - 4ac$ with parentheses around each substituted value prevents most mistakes.