Nonlinear equations in one variable: solve quadratics by factoring, the quadratic formula and completing the square, and solve radical, rational and exponential equations, checking for extraneous solutions.

What this skill is asking

A nonlinear equation in one variable has the variable raised to a power other than one, or under a root, or in an exponent. The Digital SAT (Advanced Math domain) focuses heavily on quadratics (solved by factoring, the quadratic formula, or completing the square) and also tests radical, rational, and exponential equations. A recurring theme is checking for extraneous solutions that the algebra introduces but the original equation rejects.

Solving quadratics

Three methods cover every quadratic; pick the fastest for the numbers.

A worked quadratic-formula solve

When factoring fails, the formula is reliable.

Radical and rational equations

For a radical equation, isolate the root and square both sides, which can introduce a value that does not satisfy the original equation, so you must check each candidate. For example, $\sqrt{x} = x - 6$ squares to $x = x^2 - 12x + 36$, giving $x = 4$ or $x = 9$; testing shows $x = 9$ works but $x = 4$ does not ($\sqrt{4} = 2 \neq -2$), so $x = 4$ is extraneous. For a rational equation, multiply through by the common denominator to clear fractions, solve the resulting polynomial, and discard any solution that makes a denominator zero.

Exponential equations

When the variable is in the exponent, the key move is to write both sides with the same base and then equate the exponents, because $b^m = b^n$ implies $m = n$. For example, $2^{x+1} = 8$ becomes $2^{x+1} = 2^3$, so $x + 1 = 3$ and $x = 2$. If the bases cannot be matched easily, the SAT version is usually solvable by recognising a power (rewrite $9$ as $3^2$, $\frac{1}{4}$ as $2^{-2}$) or by graphing in Desmos and reading the intersection. Exponential equations connect to growth and decay models, where the same same-base technique solves for a time or a rate.

Quadratics in disguise and the zero-product property

Some SAT equations are not obviously quadratic but become so after a substitution or a rearrangement. An equation like $x^4 - 5x^2 + 4 = 0$ is a quadratic in $x^2$: let $u = x^2$, solve $u^2 - 5u + 4 = (u - 1)(u - 4) = 0$ to get $u = 1$ or $u = 4$, then back-substitute to find $x = \pm 1$ and $x = \pm 2$. The deeper principle behind all factored solving is the zero-product property: if a product equals zero, at least one factor is zero. That is why setting $(x - 7)(x + 2) = 0$ lets you write $x - 7 = 0$ or $x + 2 = 0$. The property only works when one side is zero, so always move everything to one side and set the equation equal to zero before factoring; trying to factor against a nonzero right-hand side is a common and costly mistake.