PSAT Advanced Math — radical, rational and absolute value equations; a line meeting a parabola

The Advanced Math domain of the PSAT/NMSQT goes beyond quadratics: its skill list names linear absolute value equations, simple rational and radical equations in one variable, and systems made of a linear and a nonlinear equation. These questions reward careful checking more than long computation, because the usual methods — squaring both sides, clearing denominators, splitting an absolute value into cases — can produce answers that do not work in the original equation, or miss one that does.

This material trains those three equation types and the line–parabola system. For radical equations you isolate the root, square, and check each candidate, so that you catch an extraneous solution and recognize an equation such as a square root equal to a negative number that has no solution at all. For rational equations you clear the denominators and reject any value that would make one of them zero — including an equation whose only candidate is excluded. For absolute value equations you split the equation into its two cases and recognize when the right side allows two solutions, one or none.

For a system of a line and a parabola you substitute the linear expression into the quadratic, solve, and read the number of intersection points from the discriminant: two crossings, one point of tangency, or none. You find the constant that makes the line touch the parabola exactly once, and one question shows both graphs on a grid so you can connect the algebraic solutions with the points where the graphs cross. A context question uses a square-root model for the time an object takes to fall.

The quiz has twelve multiple-choice questions with explanations that show where each wrong option comes from. The flashcards summarize the checks to make and the three outcomes of a line–parabola system. The printable written work asks you to solve, check and justify, which is what the student-produced response questions of the test require. The oral exam lets you solve one equation at a time with an examiner who asks you to explain every check.

The content is based on the College Board's published skill descriptions for the PSAT/NMSQT Math section (Advanced Math: nonlinear equations in one variable and systems of equations in two variables). It is independent practice, not produced or endorsed by the College Board.

  • Solve simple radical equations and reject extraneous solutions
  • Solve simple rational equations and exclude values that make a denominator zero
  • Solve linear absolute value equations and tell when they have two, one or no solutions
  • Solve a system of a linear and a quadratic equation and use the discriminant to count its solutions
  • Relate the solutions of a line–parabola system to the intersection points of their graphs

Practice material written by Zestly, based on the College Board's skill descriptions for the digital PSAT/NMSQT Math section (Advanced Math: nonlinear equations in one variable — linear absolute value, simple rational and radical equations; systems of linear and nonlinear equations in two variables and their graphs), as published on satsuite.collegeboard.org and in the Assessment Framework for the Digital SAT Suite.

Sample question

A coordinate grid showing the parabola y = −x² + 6x − 5, which opens downward, and the line y = x − 1; the two graphs cross at two points.

The graph shows the parabola $y = -x^2 + 6x - 5$ and the line $y = x - 1$. The system formed by the two equations has two solutions, $(x_1, y_1)$ and $(x_2, y_2)$. What is the value of $y_1 + y_2$?

See the answer

$3$

Setting $-x^2 + 6x - 5 = x - 1$ gives $x^2 - 5x + 4 = 0$, so $x = 1$ or $x = 4$. On the line, $y = 0$ when $x = 1$ and $y = 3$ when $x = 4$: the intersection points are $(1, 0)$ and $(4, 3)$, as the graph shows. So $y_1 + y_2 = 0 + 3 = 3$. The value $5$ is $x_1 + x_2$, and $4$ is the $y$-coordinate of the vertex of the parabola.

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