# Hard SAT Math Worksheet PDF with Answers | 1600.now

Source: https://1600.now/sat-hard-math-worksheet

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## Hard SAT Math Worksheet

For hard SAT math practice, identify the requested quantity and the equation’s structure before doing a long calculation. Root relationships, parameter conditions, and domain restrictions can shorten the work and catch a false solution.

Written by [Luke Finigan](https://1600.now/about)

This original 12-question set covers challenging algebra, exponential models, and successive percent changes. It is a topic practice worksheet, not a scored adaptive test. Download it with or without the explained answer key.

12 questions · about 30 minutes · Math › mixed skills

[Download 6-page PDF](https://1600.now/downloads/worksheets/sat-hard-math-worksheet-student-letter.pdf)

[Share to Google Classroom](https://classroom.google.com/share?url=https%3A%2F%2F1600.now%2Fsat-hard-math-worksheet) · This worksheet is free to print and share under [CC BY-NC-ND 4.0](https://creativecommons.org/licenses/by-nc-nd/4.0/). Tutors may also use it in paid sessions. If you post it online, please link to this page instead of re-uploading the PDF. These are original questions, not College Board items.

### Look for a relationship before expanding

For $ax^2+bx+c=0$, the sum of the roots is $-\dfrac ba$ and their product is $\dfrac ca$. If the roots have a stated ratio or difference, represent them with one variable, then use the sum or product to determine it.

Exactly one distinct real quadratic solution requires $b^2-4ac=0$ when $a\ne0$. A parameter appearing inside $c$ is not necessarily the answer itself; solve the resulting parameter equation and then reconstruct the quadratic to check.

For exponential functions $f(x)=ab^x$ with $b>0$, a ratio of outputs separates the growth factor from the starting value: $\dfrac{f(v)}{f(u)}=b^{v-u}$. Count the interval length before taking a root or stepping backward.

After any transformation, return to the original restrictions and requested expression. A radical candidate must satisfy the unsquared equation, and squaring a sum must include its middle term.

Try it yourself

A monic quadratic has positive roots $2$ and $7$. What is the coefficient $k$ in $x^2-kx+14=0$?

A $5$ B $9$ C $14$

### Questions

1. **Question 1**

   The equation $x^2 - kx + 48 = 0$ has two positive solutions. One solution is three times the other. What is $k$?

   A. 8 B. 16 C. 12 D. 48

   **B.** Write the roots as $r$ and $3r$. Their product is 48, so $3r^2 = 48$ and $r = 4$ because both roots are positive. The roots are 4 and 12. Their sum equals $k$, so $k = 16$.
2. **Question 2**

   The square root of $2x + 9$ is equal to $x - 3$. What value of $x$ satisfies the equation?

   A. 8 B. 3 C. 6 D. 0

   **A.** Because a square root is nonnegative, $x$ must be at least 3. Squaring gives $2x + 9 = x^2 - 6x + 9$, so $x(x - 8) = 0$. The candidate 0 violates $x \ge 3$. The valid solution is 8: the square root of 25 equals $8 - 3$.
3. **Question 3**

   The graph $y=ax^{2}+bx+c$ has vertex (3, −4) and passes through (0, 14). What is $a+b+c$?

   A. 2 B. 12 C. 4 D. 14

   **C.** Vertex form is $y=a(x-3)^{2}-4$. Using (0, 14) gives $14=9a-4$, so $a=2$. Expanding gives $y=2x^{2}-12x+14$. Thus $a+b+c=2-12+14=4$. Equivalently, evaluate $y$ at $x=1$.
4. **Question 4**

   The line $y=mx+7$ intersects the parabola $y=x^{2}+3x+11$ at exactly one point. If $m$ is positive, what is $m$?

   A. 4 B. 1 C. 3 D. 7

   **D.** Equating the expressions for $y$ gives $x^{2}+(3-m)x+4=0$. A single intersection requires $(3-m)^{2}-16=0$. Thus $3-m$ is 4 or −4, giving $m=-1$ or 7. The positive value is 7.
5. **Question 5**

   A positive quantity is decreased by $p\%$, then the result is increased by $p\%$, where $0<p<100$. The final quantity is 84% of the original. What is $p$?

   A. 16 B. 40 C. 32 D. 84

   **B.** The combined multiplier is $(1-\dfrac{p}{100})(1+\dfrac{p}{100})=1-\dfrac{p^{2}}{10000}$. Set this equal to 0.84: $\dfrac{p^{2}}{10000}=0.16$, so $p^{2}=1600$. Since $p$ is positive, $p=40$. Checking: 0.60 times 1.40 equals 0.84.
6. **Question 6**

   For the quadratic function $f(x)=ax^{2}+bx+c$, $f(1)=8$, $f(3)=26$, and $f(5)=60$. What is $f(8)$?

   A. 120 B. 133 C. 141 D. 136

   **C.** The $x$-values are spaced by 2. Subtracting the first two equations gives $8a+2b=18$; subtracting the next two gives $16a+2b=34$. Hence $8a=16$, so $a=2$ and $b=1$. From $f(1)=8$, $c=5$. Therefore $f(8)=2(64)+8+5=141$.
7. **Question 7**

   The equation $x^{2}-10x+k=0$ has two positive roots whose difference is 4. What is $k$?

   A. 9 B. 16 C. 21 D. 24

   **C.** The roots sum to 10 and differ by 4, so they are 3 and 7. Their product equals the constant $k$, giving $k=21$.
8. **Question 8**

   For positive constants $a$ and $b$, $f(x)=a(b^{x})$. If $f(2)=12$ and $f(5)=96$, what is $f(0)$?

   A. 2 B. 3 C. 4 D. 6

   **B.** Dividing gives $b^{3}=\dfrac{96}{12}=8$, so $b=2$. Then $4a=12$, hence $a=3$ and $f(0)=3$.
9. **Question 9**

   A quadratic $f$ has zeros −2 and 6. Its minimum value is −32. What is $f(0)$?

   A. −32 B. −24 C. −16 D. 24

   **B.** Write $f(x)=a(x+2)(x-6)$. The vertex lies midway between the zeros at $x=2$, where $-16a=-32$, so $a=2$. Then $f(0)=2(2)(-6)=-24$.
10. **Question 10**

    For which positive value of $k$ does $x^{2}-6x+k=2x-k$ have exactly one real solution?

    A. 4 B. 8 C. 12 D. 16

    **B.** Rearrange to $x^{2}-8x+2k=0$. A repeated root requires discriminant $64-8k=0$, so $k=8$.
11. **Question 11**

    A positive quantity is increased by $p\%$ and then increased by $2p\%$. The result is 168% of the original. What is $p$?

    A. 10 B. 20 C. 30 D. 40

    **B.** Set $(1+\dfrac{p}{100})(1+\dfrac{2p}{100})=1.68$. With $p=20$ the factors are 1.2 and 1.4, whose product is 1.68. The other algebraic root is negative and is excluded.
12. **Question 12**

    If $x+\dfrac{1}{x}=5$ for a nonzero real $x$, what is $x^{2}+\dfrac{1}{x^{2}}$?

    A. 15 B. 21 C. 23 D. 25

    **C.** Squaring gives $x^{2}+2+\dfrac{1}{x^{2}}=25$. Subtract 2 to obtain the requested expression, 23.

0/12 correct

### Check your hard math answers

Each row is a wrong answer choice from one of the sheet's problems. Find the one you picked, then run the check in the last column.

| What your answer looked like | What actually happened | Fix it next time |
| --- | --- | --- |
| 0 in problem 2 | Squaring both sides added a solution. At $x = 0$, $\sqrt{9} = 3$, but $x - 3 = -3$. | A square root can't be negative, so $x - 3 \ge 0$ and $x \ge 3$. Only 8 works: $\sqrt{25} = 5 = 8 - 3$. |
| 16 in problem 5 | 16 is $100 - 84$, the net loss. Each change has to be larger than that, because the increase is taken from a smaller amount. Test 16: $0.84 \times 1.16 \approx 0.974$, not 0.84. | Multiply the factors: $(1 - \tfrac{p}{100})(1 + \tfrac{p}{100}) = 1 - \tfrac{p^2}{10{,}000} = 0.84$, so $p^2 = 1{,}600$ and $p = 40$. Check: $0.60 \times 1.40 = 0.84$. |
| 24 in problem 7 | 24 is $4 \times 6$. Those roots add to 10, but they differ by 2, not 4. | Find two numbers that add to 10 and differ by 4: 3 and 7. In $x^2 - 10x + k$, $k$ is their product, 21. Check: $(x - 3)(x - 7) = x^2 - 10x + 21$. |
| 6 in problem 8 | 6 is $f(1)$: you divided $f(2) = 12$ by $b$ once. Going from $x = 2$ back to $x = 0$ divides by $b$ twice. | From $\dfrac{f(5)}{f(2)} = b^3 = 8$, $b = 2$. Then $f(0) = \dfrac{12}{2^2} = 3$. |
| 16 in problem 10 | 16 is the value of $2k$, the constant term of $x^2 - 8x + 2k = 0$. | One solution means the discriminant is 0: $64 - 4(2k) = 0$, so $8k = 64$ and $k = 8$. Check: $x^2 - 8x + 16 = (x - 4)^2$. |
| 25 in problem 12 | You squared each term and dropped the middle one. $(x + \tfrac{1}{x})^2 = x^2 + 2 + \tfrac{1}{x^2}$, because $2 \cdot x \cdot \tfrac{1}{x} = 2$. | Square both sides: $x^2 + 2 + \tfrac{1}{x^2} = 25$. Subtract 2 to get 23. |

#### Worked example: roots in a 1-to-3 ratio

Problem 1 says $x^2 - kx + 48 = 0$ has two positive solutions, one three times the other. Call them $r$ and $3r$. Then the quadratic factors as $(x - r)(x - 3r) = x^2 - 4rx + 3r^2$.

Match the constants: $3r^2 = 48$, so $r^2 = 16$ and $r = 4$, since the roots are positive. The roots are 4 and 12. Match the $x$-terms: $k = 4r = 16$. Check: $x^2 - 16x + 48 = (x - 4)(x - 12)$. The choice 12 is the larger root, and 48 is the product of the roots.

### Apply the check to fresh questions

For each miss, name the structure you could have used: root sum/product, discriminant, growth factor, or domain restriction. Use the Math practice link to apply it without a worksheet topic cue.

[Practice Math questions](https://1600.now/bank/math/browse)

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