Nonlinear equations and systems

Core ideas

Why extraneous solutions happen

Two moves can create fake solutions that fail the original equation:

• Squaring both sides. Squaring cannot tell $3 = 3$ from $3 = -3$.
• Multiplying by an expression containing $x$. A value that makes a denominator 0 can survive the algebra.

So every radical and rational equation ends with a check. On a multiple-choice question, the trap choice is almost always “both candidates”.

Radical equations

1. Isolate the radical.
2. Square both sides (square the whole other side: $(x + 2)^2 = x^2 + 4x + 4$).
3. Solve the resulting equation.
4. Substitute each candidate into the original. A square root is never negative, so any candidate that makes the non-radical side negative is extraneous.

Rational equations

1. List the excluded values, where any denominator equals 0.
2. Multiply every term, including whole-number terms, by the least common denominator.
3. Solve, then discard any excluded value.

A rational equation with a parameter has no solution when its only candidate equals an excluded value, or when the equation collapses to a false statement such as $2 = 1$.

Absolute value equations

$|A| = B$ splits into $A = B$ or $A = -B$. If $B$ is a negative number, there is no solution. If $B$ contains $x$, both candidates must make $B \geq 0$, so check them.

Exponential equations with a common base

Rewrite both sides as powers of the same base, then set the exponents equal. Know these powers on sight:

Also, $\frac{1}{8} = 2^{-3}$ and $\sqrt{2} = 2^{\frac{1}{2}}$. Distribute carefully: $8^{x-1} = 2^{3(x - 1)} = 2^{3x - 3}$.

Nonlinear systems

Substitute one equation into the other to get a single equation in $x$, solve it, then find $y$ from the simpler original equation. The number of solutions is the number of intersection points:

Worked examples

Exam technique

• Let Desmos do the checking. Graph the left side and the right side as two separate equations. The intersections are the true solutions, and extraneous roots never appear. This is the fastest defense against the “both candidates” trap.
• Systems: type both equations into Desmos and click the intersection points. For a condition such as $a > 0$, pick the point on the correct side of the axis.
• Parameters: enter the equations with the letter and add a slider. Tangency (exactly one solution) is where the graphs just touch. Then confirm with $D = 0$ algebraically, because the slider only gets you close.
• Horizontal-line shortcut: for $y = k$ against a parabola, compare $k$ with the vertex’s $y$-value instead of computing a discriminant.
• Two parabolas: subtract the equations. The $x$-coordinates of the intersections are the roots of the difference. Then get the $y$-values from either original equation.
• Budget: hard parameter items take 90–120 seconds. If you’re past 2 minutes, make your best guess, flag it, and return with the time you’ve saved on easier items.

Common mistakes

Quick recap

• Squaring both sides or clearing denominators can create extraneous solutions. Always check candidates in the original equation.
• Radical: isolate, square, solve, check that the non-radical side is not negative.
• Rational: list excluded values first, multiply every term, discard excluded values.
• Absolute value: two cases, and the right side must be nonnegative.
• Exponential: use a common base and distribute into the exponent.
• Line and parabola: substitute, then read the discriminant (two, one or zero intersections). Tangency means $D = 0$.
• Desmos graphs of both sides show only the true solutions.