SAT practice test

SAT Advanced Math Practice Test

Quadratics, exponentials, equivalent expressions, nonlinear functions. 35% of Math. Here are 10 SAT advanced math questions, each with the answer and the reason.

Updated 2026-11-14 by the Tigertest team. Original questions written to the digital SAT specifications. Not affiliated with College Board. Score estimates are unofficial.

1. 1.

   Which of the following is equivalent to $9x^2 - 25$?

   • A.

     $(3x - 5)(3x + 5)$

   • B.

     $(3x - 5)^2$

   • C.

     $(9x - 5)(x + 5)$

   • D.

     $(3x + 5)^2$

   Answer: A.

   The expression is a difference of squares: $9x^2 = (3x)^2$ and $25 = 5^2$, and $a^2 - b^2 = (a - b)(a + b)$, so $9x^2 - 25 = (3x - 5)(3x + 5)$. Expanding confirms it: the middle terms $15x$ and $-15x$ cancel. The choice $(3x - 5)^2$ expands to $9x^2 - 30x + 25$, which has a middle term and the wrong sign on $25$. The choice $(9x - 5)(x + 5)$ expands to $9x^2 + 40x - 25$.

2. 2.

   What are the solutions to the equation $2x^2 - 6x + 1 = 0$?

   • A.

     $x = \frac{3 \pm \sqrt{7}}{2}$

   • B.

     $x = \frac{-3 \pm \sqrt{7}}{2}$

   • C.

     $x = \frac{3 \pm \sqrt{11}}{2}$

   • D.

     $x = \frac{6 \pm \sqrt{7}}{2}$

   Answer: A.

   With $a = 2$, $b = -6$ and $c = 1$, the quadratic formula gives $x = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4}$. Since $\sqrt{28} = 2\sqrt{7}$, this simplifies to $\frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2}$. The choice $\frac{-3 \pm \sqrt{7}}{2}$ uses $+b$ instead of $-b$ in the numerator. The choice $\frac{3 \pm \sqrt{11}}{2}$ adds $4ac$ instead of subtracting it ($36 + 8 = 44$), and $\frac{6 \pm \sqrt{7}}{2}$ divides only the radical by $2$ when simplifying.

3. 3.

   In the equation $kx^2 - 6x + 3 = 0$, $k$ is a nonzero constant. If the equation has exactly one real solution, what is the value of $k$?

   • A.

     $-3$

   • B.

     $3$

   • C.

     $9$

   • D.

     $12$

   Answer: B.

   A quadratic $ax^2 + bx + c = 0$ has exactly one real solution when its discriminant $b^2 - 4ac$ equals $0$. Here $b^2 - 4ac = (-6)^2 - 4(k)(3) = 36 - 12k$, and $36 - 12k = 0$ gives $k = 3$. Check: $3x^2 - 6x + 3 = 3(x - 1)^2$, which is $0$ only at $x = 1$. The choice $12$ comes from leaving out the $4$ in $4ac$ ($36 - 3k = 0$). The choice $-3$ comes from treating $b^2$ as $-36$, and $9$ comes from setting $b^2 = 4k$, dropping $c$.

4. 4.

   Selected values of the exponential function $f$ are $f(0) = 4$, $f(1) = 6$, $f(2) = 9$, and $f(3) = 13.5$. Which equation defines $f$?

   • A.

     $f(x) = 4\left(\frac{3}{2}\right)^x$

   • B.

     $f(x) = \frac{3}{2}(4)^x$

   • C.

     $f(x) = 4\left(\frac{2}{3}\right)^x$

   • D.

     $f(x) = 4 + 2x$

   Answer: A.

   An exponential function has the form $f(x) = a \cdot b^x$, where $a = f(0)$ and $b$ is the ratio of consecutive outputs. Here $a = 4$, and $\frac{6}{4} = \frac{9}{6} = \frac{13.5}{9} = 1.5 = \frac{3}{2}$, so $f(x) = 4\left(\frac{3}{2}\right)^x$. The choice $f(x) = 4 + 2x$ fits the first two values but not $f(2) = 9$; the outputs grow by a constant factor, not a constant difference. The choice $\frac{3}{2}(4)^x$ swaps the starting value and the growth factor.

5. 5.

   The function $g$ is defined by $g(x) = f(x - 4) + 1$. If the point $(2, 5)$ lies on the graph of $y = f(x)$, which point must lie on the graph of $y = g(x)$?

   • A.

     $(6, 6)$

   • B.

     $(-2, 6)$

   • C.

     $(6, 4)$

   • D.

     $(-2, 4)$

   Answer: A.

   Replacing $x$ with $x - 4$ shifts the graph of $f$ to the right by $4$, and adding $1$ shifts it up by $1$. The point $(2, 5)$ moves to $(2 + 4, 5 + 1) = (6, 6)$. Check: $g(6) = f(6 - 4) + 1 = f(2) + 1 = 6$. The choice $(-2, 6)$ shifts left, the most common error, because $x - 4$ looks like it should move the graph in the negative direction; it does not, since the input must be $4$ larger to reach the same value of $f$.

6. 6.

   Which of the following expressions is equivalent to $3(2x - 5) + 4x$?

   • A.

     $6x - 15$

   • B.

     $10x - 15$

   • C.

     $10x - 5$

   • D.

     $10x + 15$

   Answer: B.

   Distribute the $3$ to both terms inside the parentheses: $3(2x - 5) = 6x - 15$. Then combine like terms: $6x - 15 + 4x = 10x - 15$. The choice $10x - 5$ comes from multiplying $3$ by $2x$ but forgetting to multiply $3$ by $-5$, and $6x - 15$ comes from dropping the $4x$ term.

7. 7.

   A bacteria population starts at $500$ cells and doubles every $3$ hours. Which of the following functions gives the population $P(t)$, in cells, $t$ hours after the start?

   • A.

     $P(t) = 500 \cdot 2^{t/3}$

   • B.

     $P(t) = 500 \cdot 2^{3t}$

   • C.

     $P(t) = 500 \cdot 3^{t/2}$

   • D.

     $P(t) = 500 + \frac{1000}{3}t$

   Answer: A.

   The population starts at $500$ and is multiplied by $2$ once every $3$ hours, so after $t$ hours it has doubled $\frac{t}{3}$ times. That gives $P(t) = 500 \cdot 2^{t/3}$; as a check, $P(3) = 1000$ and $P(6) = 2000$. The choice $500 \cdot 2^{3t}$ reverses the fraction and would double the population three times every hour, and $500 + \frac{1000}{3}t$ grows by a constant amount instead of a constant factor.

8. 8.

   Which of the following expressions is equivalent to $(3x^2 - 5x + 2) - (x^2 - 4x - 6)$?

   • A.

     $2x^2 - x + 8$

   • B.

     $2x^2 - 9x - 4$

   • C.

     $2x^2 - x - 4$

   • D.

     $2x^2 - 9x + 8$

   Answer: A.

   Distribute the subtraction to every term of the second polynomial: $3x^2 - 5x + 2 - x^2 + 4x + 6$. Combining like terms gives $2x^2 - x + 8$. Choice B comes from changing the sign of only the first term in the second parentheses and then computing $-5x - 4x$ and $2 - 6$; the sign of every term inside the subtracted parentheses must change.

9. 9.

   Which of the following is equivalent to $\frac{\sqrt[3]{x^5} \cdot x^2}{\sqrt{x}}$ for all $x > 0$?

   • A.

     $x^{5/3}$

   • B.

     $x^{17/6}$

   • C.

     $x^{19/6}$

   • D.

     $x^{25/6}$

   Answer: C.

   Rewrite each radical with a rational exponent: $\sqrt[3]{x^5} = x^{5/3}$ and $\sqrt{x} = x^{1/2}$. Multiplying adds exponents and dividing subtracts them: $\frac{5}{3} + 2 - \frac{1}{2} = \frac{10}{6} + \frac{12}{6} - \frac{3}{6} = \frac{19}{6}$. Choice D adds $\frac{1}{2}$ instead of subtracting it, treating the division by $\sqrt{x}$ as multiplication. Choice B multiplies $\frac{5}{3}$ by $2$ instead of adding $2$.

10. 10.

    A public library's collection of e-books has $4{,}000$ titles today, and the library plans to increase the number of titles by $10\%$ each year. Which of the following functions gives the number of titles $N(t)$ in the collection $t$ years from today?

    • A.

      $N(t) = 4000(1.1)^t$

    • B.

      $N(t) = 4000(0.1)^t$

    • C.

      $N(t) = 4000 + 400t$

    • D.

      $N(t) = 1.1(4000)^t$

    Answer: A.

    Increasing a quantity by $10\%$ multiplies it by $1 + 0.10 = 1.1$, and this happens once each year, so after $t$ years the collection has $N(t) = 4000(1.1)^t$ titles. The choice $N(t) = 4000 + 400t$ adds $400$ titles every year, which is $10\%$ of the original collection, not $10\%$ of the current collection, so it is linear growth rather than exponential growth. The choice $N(t) = 4000(0.1)^t$ multiplies by $0.1$ each year, which would shrink the collection.