Nonlinear Equations and Systems

Factor a quadratic and use the zero-product rule

For $x^2-5x+6=0$, factor to $(x-2)(x-3)=0$. A product is zero when at least one factor is zero, so the solutions are 2 and 3. If asked for the sum, answer 5; if asked for the product, answer 6.

Move all terms to one side before using this rule. From $(x-2)(x-3)=4$, you cannot set the individual factors equal to zero.

Squaring creates candidates that need a check

For $\sqrt{x+6}=x$, the right side must be nonnegative. Squaring gives $x+6=x^2$, or $(x-3)(x+2)=0$. The candidates are 3 and $-2$, but only 3 satisfies the original equation.

At $x=-2$, the left side is 2 and the right side is $-2$. A graph or a squared equation can suggest a candidate; substitution into the unsquared equation settles whether it works.

Keep domain restrictions in rational and absolute-value equations

For $\frac{x+1}{x-2}=3$, exclude $x=2$ before multiplying. Solving gives $x+1=3x-6$, so $x=\frac{7}{2}$, which is allowed. A value making an original denominator zero must always be rejected.

For $|2x-1|=5$, solve $2x-1=5$ and $2x-1=-5$, giving 3 and $-2$. An absolute value cannot equal a negative number, so $|2x-1|=-5$ has no solution.

Substitute to solve a mixed system

For $y=x^2$ and $y=x+2$, set $x^2=x+2$. Factoring gives $(x-2)(x+1)=0$. The intersections are $(2,4)$ and $(-1,1)$ after substituting into either equation.

A system can have more than one intersection. If graphing, inspect the relevant domain and both branches rather than stopping at the first visible point.

Check your understanding

Sample nonlinear equations and systems questions

Practice this skill

1. Classify the equation before choosing a method.
2. Write restrictions before clearing denominators or squaring.
3. Check every candidate and answer the exact requested root, coordinate, sum, or product.

Print the nonlinear equations and systems worksheet and worked answers.

Then use a mixed practice module to check whether you can choose this method among other question types.