PSAT 10 Math — Advanced Algebra and Function Skills

Equivalent expressions

In algebra, you often have different-looking expressions that represent the same value for every allowed input. Two expressions are equivalent expressions if they are equal for all values of the variable(s) in their domain (all values that make the expressions defined). For example, rewriting an expression by factoring or expanding doesn’t “change the math”—it changes the form, which can make certain features easier to see.

What “equivalent” really means (and why domain matters)

When you simplify or rewrite an expression, you are usually trying to produce an equivalent expression. But equivalence has a hidden condition: both expressions must be defined for the same inputs.

• $\frac{x^2-1}{x-1}$ can be simplified algebraically, but you must remember that the original expression is undefined at $x=1$.
• If you cancel incorrectly and then treat the new expression as defined at $x=1$, you’ve changed the domain—so you no longer have a truly equivalent expression.

That’s why, on PSAT-style questions, you’ll often see wording like “for all $x \ne 1$” or “where defined.”

Core rewriting tools

Most equivalence problems come down to a few “moves.” The goal is to use algebra rules that preserve equality.

Combine like terms

Like terms have the same variable part (same variables to the same powers). Combining like terms is basically “counting” them.

If you have

$3x^2+5x-2x^2+7$

you can combine $3x^2-2x^2$ because both are $x^2$ terms.

Distribute and factor (opposite directions)

The distributive property expands products:

$a(b+c)=ab+ac$

Factoring reverses distribution:

$ab+ac=a(b+c)$

These are essential because different forms reveal different information:

• Expanded form helps you combine like terms and compare coefficients.
• Factored form helps you see zeros (solutions where the expression equals zero) and solve equations.

Exponent rules (carefully)

Exponent rules are frequent in Advanced Math because they allow you to rewrite expressions efficiently.

Key rules (for real-number bases where the expressions are defined):

$a^m \times a^n=a^{m+n}$

$\frac{a^m}{a^n}=a^{m-n}$

$(a^m)^n=a^{mn}$

$a^{-n}=\frac{1}{a^n}$

A common trap is assuming these rules apply when you have addition:

$(a+b)^2 \ne a^2+b^2$

Instead:

$(a+b)^2=a^2+2ab+b^2$

Radicals and rational exponents

A square root is a power:

$\sqrt{x}=x^{\frac{1}{2}}$

More generally:

$\sqrt[n]{x}=x^{\frac{1}{n}}$

This helps you rewrite expressions into a form that is easier to simplify or solve. But note the domain issue: over the real numbers, $\sqrt{x}$ requires $x\ge 0$.

Equivalent forms for quadratics

Quadratic expressions appear constantly in PSAT Advanced Math. You should recognize three common equivalent forms, each useful for different tasks.

Standard form

$ax^2+bx+c$

• Easy to read the $y$-intercept $c$ when the expression is a function $f(x)=ax^2+bx+c$.

Factored form

$a(x-r_1)(x-r_2)$

• Easy to find zeros: $x=r_1$ and $x=r_2$.

Vertex form

$a(x-h)^2+k$

• Easy to find the vertex $(h,k)$ and the maximum/minimum value.

You can move between these forms by expanding, factoring, or completing the square.

Worked examples (equivalent expressions)

Example 1: Factoring to create an equivalent expression

Rewrite

$x^2-9x$

in factored form.

Step 1: Factor out the greatest common factor. Both terms share $x$.

$x^2-9x=x(x-9)$

This is equivalent for all real $x$.

Example 2: Simplifying a rational expression (and tracking restrictions)

Simplify

$\frac{x^2-4}{x-2}$

Step 1: Factor the numerator (difference of squares).

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

Step 2: Cancel the common factor.

$\frac{(x-2)(x+2)}{x-2}=x+2$

Step 3: State the restriction. The original denominator requires

$x-2\ne 0 \Rightarrow x\ne 2$

So the simplified expression is

$x+2 \text{ for } x\ne 2$

If a multiple-choice option says just $x+2$, the question often implies “for all $x\ne 2$” or “where defined.”

Example 3: Completing the square to rewrite a quadratic

Rewrite

$x^2+6x+1$

in vertex form.

Step 1: Group the quadratic and linear terms.

$x^2+6x+1=(x^2+6x)+1$

Step 2: Add and subtract the number that completes the square. Take half of $6$ (which is $3$), square it (which is $9$).

$x^2+6x+1=(x^2+6x+9)-9+1$

Step 3: Factor the perfect square trinomial.

$x^2+6x+9=(x+3)^2$

So:

$x^2+6x+1=(x+3)^2-8$

Now you can see the vertex quickly if this is $f(x)$: the vertex is $(-3,-8)$.

Exam Focus

• Typical question patterns:
  • “Which expression is equivalent to … ?” often requiring factoring, distributing, or exponent rules.
  • Simplify a rational expression and choose the matching option (sometimes with domain restrictions).
  • Rewrite a quadratic into vertex or factored form to identify a feature (vertex, zeros, max/min).
• Common mistakes:
  • Cancelling across addition, such as $\frac{x+2}{x}=2$ (not valid).
  • Forgetting domain restrictions after simplifying rational or radical expressions.
  • Misusing exponent rules, especially assuming $(a+b)^n=a^n+b^n$.

Nonlinear equations in one variable and systems of equations in two variables

A nonlinear equation is an equation where the variable is not only to the first power and not just in a simple linear form. The graph of a nonlinear equation is not a straight line. On the PSAT, the most common nonlinear equations involve quadratics, absolute value, radicals, rational expressions, and occasionally exponentials.

Why this matters: linear equations have one “main” solving pattern, but nonlinear equations can have multiple solutions, no real solutions, or solutions that must be checked because algebra steps can introduce invalid answers.

Solving nonlinear equations in one variable

Quadratic equations

A quadratic equation can often be written as

$ax^2+bx+c=0$

You typically solve quadratics by:

1. Factoring (fastest when it works)
2. Square root method (best when you have $(x-h)^2=k$)
3. Quadratic formula (works for all quadratics, but more computation)

Quadratic formula:

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

The expression under the square root,

$b^2-4ac$

is the discriminant. It tells you how many real solutions exist:

• If $b^2-4ac>0$, two real solutions
• If $b^2-4ac=0$, one real solution (a repeated root)
• If $b^2-4ac<0$, no real solutions

Absolute value equations

Absolute value measures distance from zero. So

$|x|=k$

means $x$ is $k$ units away from zero. If $k>0$, there are two solutions:

$x=k \quad \text{or} \quad x=-k$

For equations like

$|x-a|=k$

you’re measuring distance from $a$.

A reliable method is to split into two cases:

$x-a=k \quad \text{or} \quad x-a=-k$

A major “what goes wrong” moment: if $k<0$, there is no solution because distance cannot be negative.

Radical equations (square roots)

When variables appear under a radical, like

$\sqrt{x+1}=x-1$

you often solve by squaring both sides. Squaring can create extra solutions because

$(-2)^2=2^2$

So you must check solutions in the original equation.

Also keep domain constraints in mind:

• The radicand must be nonnegative: $x+1\ge 0$.
• If you have $\sqrt{\text{something}}=x-1$, the right side must also be nonnegative because a square root is never negative in real numbers.

Rational equations (variables in denominators)

Equations like

$\frac{1}{x-3}=2$

require you to exclude values that make denominators zero, here $x\ne 3$. Multiplying both sides by a common denominator is typical, but keep the restriction and check at the end.

Systems involving nonlinear equations

A system of equations asks for values that satisfy both equations simultaneously. In two variables, solutions correspond to intersection point(s) of graphs.

Nonlinear systems on PSAT often include:

• A line and a parabola
• Two parabolas
• A circle and a line (sometimes)

You’ll usually use one of two strategies:

1. Substitution: solve one equation for a variable and substitute into the other.
2. Elimination: less common with nonlinear systems, but sometimes works if terms cancel.

Graph reasoning is also important: you might not fully solve, but you may be asked how many solutions exist by thinking about intersections.

Worked examples (nonlinear equations and systems)

Example 1: Solving a quadratic by factoring

Solve:

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

Step 1: Factor. Find two numbers that multiply to $6$ and add to $-5$: $-2$ and $-3$.

$x^2-5x+6=(x-2)(x-3)$

Step 2: Use the zero-product property. If a product is zero, at least one factor is zero.

$x-2=0 \Rightarrow x=2$

$x-3=0 \Rightarrow x=3$

So the solutions are $2$ and $3$.

Example 2: Radical equation with checking

Solve:

$\sqrt{x+5}=x-1$

Step 1: Apply domain logic first.

• Need $x+5\ge 0 \Rightarrow x\ge -5$.
• Also need $x-1\ge 0 \Rightarrow x\ge 1$.

So any solution must satisfy $x\ge 1$.

Step 2: Square both sides.

$(\sqrt{x+5})^2=(x-1)^2$

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

Step 3: Rearrange into a quadratic.

$0=x^2-3x-4$

Step 4: Factor.

$x^2-3x-4=(x-4)(x+1)$

So candidates are $x=4$ and $x=-1$.

Step 5: Check candidates in the original equation (and domain).

• $x=4$: $\sqrt{9}=3$ and $4-1=3$, works.
• $x=-1$: fails domain $x\ge 1$, also $\sqrt{4}=2$ but $-1-1=-2$, not equal.

Solution:

$4$

Example 3: System of a line and a parabola (substitution)

Solve the system:

$y=x^2$

$y=2x+3$

Step 1: Substitute because both equal $y$.

$x^2=2x+3$

Step 2: Rearrange into quadratic form.

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

Step 3: Factor.

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

So $x=3$ or $x=-1$.

Step 4: Find $y$ for each value. Using $y=x^2$:

• If $x=3$, then $y=9$.
• If $x=-1$, then $y=1$.

Solutions (intersection points):

$(3,9)$

$(-1,1)$

Notice how a line can intersect a parabola in 0, 1, or 2 points—this is why nonlinear systems can have multiple solutions.

Exam Focus

• Typical question patterns:
  • Solve a quadratic or radical equation and select the correct solution set.
  • Solve a system like “line + parabola” via substitution; sometimes the question only asks for the number of solutions.
  • Interpret solutions as intersection points on a graph.
• Common mistakes:
  • Forgetting to check for extraneous solutions after squaring both sides in radical equations.
  • Losing or adding solutions when factoring incorrectly or mishandling the zero-product property.
  • Ignoring restrictions like denominators equal to zero or requiring radicands to be nonnegative.

Nonlinear functions

A function assigns each input exactly one output. A nonlinear function is a function whose graph is not a straight line. Nonlinear functions are central to PSAT Advanced Math because they model growth, area, physics, finance, and many “real” relationships more accurately than linear ones.

Instead of only solving equations, you’ll often be asked to analyze a nonlinear function: its intercepts, vertex/turning point, end behavior, or how its graph changes when you transform it.

Function notation and key vocabulary

Function notation uses $f(x)$ to mean “the output of function $f$ when the input is $x$.” If the function is

$f(x)=x^2-4x+1$

then $f(3)$ means substitute $x=3$.

Two essential ideas:

• Domain: allowed inputs.
• Range: resulting outputs.

Domain is especially important for nonlinear functions that involve denominators and even roots.

Common nonlinear function families

Quadratic functions

A quadratic function has the form

$f(x)=ax^2+bx+c$

Its graph is a parabola.

• If $a>0$, it opens upward and has a minimum.
• If $a<0$, it opens downward and has a maximum.

The vertex is the turning point. Vertex form

$f(x)=a(x-h)^2+k$

shows that the vertex is $(h,k)$.

Zeros (or x-intercepts) solve $f(x)=0$. Factored form

$f(x)=a(x-r_1)(x-r_2)$

shows zeros at $x=r_1$ and $x=r_2$.

Exponential functions

An exponential function has the variable in the exponent, like

$f(x)=a\,b^x$

with $b>0$ and $b\ne 1$.

• If $b>1$, you have exponential growth.
• If $0<b<1$, you have exponential decay.

A key feature: exponential functions change by a constant factor for equal changes in $x$. That’s different from linear functions (constant difference) and quadratics (constant second difference).

Rational functions

A rational function is a ratio of polynomials, such as

$f(x)=\frac{x+1}{x-2}$

Rational functions often have:

• Domain restrictions where the denominator is zero.
• Vertical asymptotes where the function values grow without bound near a restricted input.

Even if asymptotes aren’t emphasized heavily, you should be comfortable with “this value is not allowed” and what that means for solutions and graphs.

Radical functions

A radical function includes a variable under a radical, like

$f(x)=\sqrt{x-3}$

Its domain requires

$x-3\ge 0 \Rightarrow x\ge 3$

Radical functions commonly appear in questions about domain, solving, and interpreting constraints.

Transformations: how changing an equation changes the graph

Transformations are a powerful way to understand a function without re-plotting everything from scratch.

If you start with a “parent” function (like $y=x^2$) and build a new function,

$y=a(x-h)^2+k$

you can interpret:

• $h$: horizontal shift (right if $h>0$, left if $h<0$)
• $k$: vertical shift (up if $k>0$, down if $k<0$)
• $a$: vertical stretch/compression and reflection (if $a<0$, reflect over the $x$-axis)

A common misconception is to reverse the direction of the horizontal shift. Because the shift is applied inside parentheses, it behaves “oppositely” from what you might guess.

For example:

$y=(x-3)^2$

is shifted right 3 (not left 3).

Interpreting nonlinear functions in context

PSAT problems often connect functions to real situations. Two patterns show up a lot:

1. Maximum/minimum from a quadratic model: For area, revenue, projectile height, or any “goes up then down” quantity, the vertex gives the peak or lowest point.
2. Growth/decay from an exponential model: Population, value over time, or repeated percent change.

In context, units matter. If $x$ is time in months, then the vertex’s $x$-coordinate is a time, and the $y$-coordinate is the quantity at that time.

Worked examples (nonlinear functions)

Example 1: Evaluating and interpreting a quadratic function

Let

$f(x)=x^2-4x+1$

Find $f(5)$.

Step 1: Substitute $x=5$.

$f(5)=5^2-4(5)+1$

Step 2: Compute.

$f(5)=25-20+1=6$

So:

$f(5)=6$

Interpretation: when the input is $5$, the output is $6$.

Example 2: Finding a vertex by completing the square

For

$g(x)=x^2+6x+2$

find the vertex.

Step 1: Complete the square for the quadratic and linear terms.

$x^2+6x+2=(x^2+6x+9)-9+2$

Step 2: Rewrite as a square.

$g(x)=(x+3)^2-7$

Now vertex form shows the vertex is

$(-3,-7)$

If this parabola opens upward (it does, since $a=1$), the minimum value is $-7$.

Example 3: Domain of a rational and a radical function

1) For

$h(x)=\frac{2}{x-8}$

the denominator cannot be zero:

$x-8\ne 0 \Rightarrow x\ne 8$

Domain: all real numbers except $8$.

2) For

$p(x)=\sqrt{3x-12}$

require the radicand to be nonnegative:

$3x-12\ge 0$

$3x\ge 12$

$x\ge 4$

Domain: $x\ge 4$.

Example 4: Exponential growth factor interpretation

Suppose

$A(t)=500\,(1.08)^t$

where $t$ is in years.

The base $1.08$ means each year the amount is multiplied by $1.08$, which corresponds to an $8\%$ increase per year. The coefficient $500$ is the initial value at $t=0$.

Exam Focus

• Typical question patterns:
  • Given a function in one form (standard, factored, vertex), identify a feature (zeros, vertex, intercept, min/max).
  • Determine domain from a formula containing a denominator or a square root.
  • Compare nonlinear functions (growth/decay, opening direction, number of solutions to $f(x)=c$).
• Common mistakes:
  • Reversing horizontal shifts in expressions like $f(x)=(x-h)^2$.
  • Forgetting that domain restrictions can remove “apparent” solutions (especially with rational functions).
  • Confusing zeros (solutions to $f(x)=0$) with the y-intercept (value at $x=0$).