Home / Digital SAT Math Practice Questions – Advanced : Radicals and rational exponents

Digital SAT Math Practice Questions – Advanced : Radicals and rational exponents

SAT MAth Practice questions – all topics

  • Advanced Math Weightage: 35% Questions: 13-15
    • Equivalent expressions
    • Nonlinear equations in one variable and systems of equations in two variables
    • Nonlinear functions

SAT MAth and English  – full syllabus practice tests

  Question   Hard

19
$
\sqrt{14-2 x}=x-7
$

What value of \(\mathrm{x}\) satisfies the given equation?

▶️Answer/Explanation

Ans:7

To solve the equation \(\sqrt{14 – 2x} = x – 7\) for \(x\):

Square both sides to eliminate the square root:
\[
(\sqrt{14 – 2x})^2 = (x – 7)^2
\]
\[
14 – 2x = (x – 7)^2
\]

Expand the right-hand side:
\[
14 – 2x = x^2 – 14x + 49
\]

Rearrange the equation to set it to zero:
\[
x^2 – 14x + 49 + 2x – 14 = 0
\]
\[
x^2 – 12x + 35 = 0
\]

Factor the quadratic equation:
\[
(x – 5)(x – 7) = 0
\]

 Solve for \(x\):
\[
x – 5 = 0 \quad \text{or} \quad x – 7 = 0
\]
\[
x = 5 \quad \text{or} \quad x = 7
\]

Check both solutions in the original equation:

For \(x = 5\):
\[
\sqrt{14 – 2(5)} = 5 – 7
\]

\[
\sqrt{4} = -2 \quad (\text{Not true, so } x = 5 \text{ is not a solution})
\]

For \(x = 7\):

\[
\sqrt{14 – 2(7)} = 7 – 7
\]
\[
\sqrt{0} = 0 \quad (\text{True, so } x = 7 \text{ is a solution})
\]

The value of \(x\) that satisfies the equation is \(7\).

  Question   Hard

Which of the following expressions is equivalent to \({(2\sqrt{x}-\sqrt{y})}^{\frac{2}{5}}\) , where x>y and y>0?

A) \({(4x-y)}^5\)

B) \(\sqrt[5]{4x-y}\)

C) \({(4x-4 \sqrt{xy}+y)}^{\frac{1}{5}}\)

D) \(\sqrt[5]{4x-4xy+y}\)

▶️Answer/Explanation

C) \({(4x-4\sqrt{xy}+y)}^{\frac{1}{5}}\)

We need to determine the expression equivalent to \((2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}\).

1. Simplify the expression inside the parentheses:
Given: \((2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}\)

2. Examine each option:

Option A: \((4 x – y)^5\)
\[
(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]

Option B: \(\sqrt[5]{4 x – y}\)
\[
(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]

Option C: \((4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}\)
\[
(4x – 4\sqrt{xy} + y)^{\frac{1}{5}} = (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]

Option D: \(\sqrt[5]{4 x – 4xy + y}\)
\[
(4x – 4xy + y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]

Thus, the correct expression is:
\[ \boxed{(4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}} \]

 Question  Hard

Which of the following is equivalent to \((\sqrt{32})(\sqrt[5]{64})\) ?

A. \(6\left(\sqrt[7]{2^5}\right)\)
B. \(6\left(\sqrt[10]{2^7}\right)\)
c. \(8\left(\sqrt[7]{2^5}\right)\)
D. \(8\left(\sqrt[10]{2^7}\right)\)

▶️Answer/Explanation

Ans:D

To simplify \((\sqrt{32})(\sqrt[5]{64})\), we can first rewrite the numbers under the square roots as powers of 2 :
\[
\begin{aligned}
& \sqrt{32}=\sqrt{2^5}=2^{\frac{5}{2}} \\
& \sqrt[5]{64}=\sqrt[5]{2^6}=2^{\frac{6}{5}}
\end{aligned}
\]

Now, when we multiply these together, we add the exponents:
\[
(\sqrt{32})(\sqrt[5]{64})=2^{\frac{5}{2}} \cdot 2^{\frac{6}{5}}=2^{\frac{5}{2}+\frac{6}{5}}=2^{\frac{25}{10}+\frac{12}{10}}=2^{\frac{37}{10}}
\]

A. \(6\left(\sqrt[7]{2^5}\right)\)

\[6\left(\sqrt[7]{2^5}\right) = 6 \times 2^{\frac{5}{7}}\]

This expression cannot be simplified to \(2^{\frac{37}{10}}\).

B. \(6\left(\sqrt[10]{2^7}\right)\)

\[6\left(\sqrt[10]{2^7}\right) = 6 \times 2^{\frac{7}{10}}\]

This expression cannot be simplified to \(2^{\frac{37}{10}}\).

C. \(8\left(\sqrt[7]{2^5}\right)\)

\[8\left(\sqrt[7]{2^5}\right) = 8 \times 2^{\frac{5}{7}}\]

This expression cannot be simplified to \(2^{\frac{37}{10}}\).

D. \(8\left(\sqrt[10]{2^7}\right)\)

\[8\left(\sqrt[10]{2^7}\right) = 8 \times 2^{\frac{7}{10}}\]

This expression can be simplified to \(2^{\frac{37}{10}}\), since \(8 \times 2^{\frac{7}{10}} = 2^{\frac{3}{10}} \times 2^{\frac{7}{10}} = 2^{\frac{3}{10} + \frac{7}{10}} = 2^{\frac{37}{10}}\).

So, after solving each option, we see that only option D simplifies to \(2^{\frac{37}{10}}\).

 Question Hard

Which of the following expressions is equivalent to $(\sqrt{2 q}+\sqrt{2 r})^{\frac{2}{3}}$, where $q>0$ and $r<0$ ?
A) $(2 q+2 r)^3$
B) $\sqrt[3]{2 q+2 r}$
C) $\sqrt[3]{2 q+2 \sqrt{q r}+2 r}$
D) $\sqrt[3]{2 q+4 \sqrt{q r}+2 r}$

▶️Answer/Explanation

D

Question

 In 480 BC, the population of the Persian Empire was approximately 49.4 million. The population of the Persian Empire was 44% of the world population at that time. Which of the following is the best estimate of the world population in 480 BC? 

  1. 21.7 million
  2. 89.1 million
  3. 93.4 million
  4. 112.3 million
▶️Answer/Explanation

D

Question Hard

1. Which expression is equivalent to \( (x^2 y)(x^4 y^{-3}) \), where \( x, y, \) and \( z \) are positive numbers?
A) \( x^6 y^{-3} \) 
B) \( x^6 y^{-2} \) 
C) \( x^4 y^{-3} \)
D) \( x^2 y^{-2} \)

▶️Answer/Explanation

Answer: B
Use exponent rules:
When multiplying terms with the same base, add the exponents.
Simplify \( x \)-terms:
$
x^2 \cdot x^4 = x^{2+4} = x^6
$
Simplify \( y \)-terms:
$
y \cdot y^{-3} = y^{1+(-3)} = y^{-2}
$
Final expression:
$
x^6 y^{-2}
$

Question Hard

2.Which of the following is equivalent to \( \sqrt{\frac{x}{64}} \) for all \( x > 0 \)?

A) \( \frac{x^2}{8} \)
B) \( \frac{x^2}{32} \)
C) \( \frac{x^{\frac{1}{2}}}{8} \)
D) \( \frac{x^{\frac{1}{2}}}{32} \)

▶️Answer/Explanation

Answer: C
Simplify \( \sqrt{\frac{x}{64}} \):
$
\sqrt{\frac{x}{64}} = \frac{\sqrt{x}}{\sqrt{64}} = \frac{x^{\frac{1}{2}}}{8}
$

Question Hard

3. Which expression is equivalent to \( \frac{x^{\frac{5}{2}}}{\sqrt{x}} \), where \( x \neq 0 \)?

A) \( x^{\frac{3}{2}} \)
B) \( x^{\frac{5}{3}} \)
C) \( x^{\frac{3}{2}} \)
D) \( x^2 \)

▶️Answer/Explanation

Answer: D
Simplify:
$
\frac{x^{\frac{5}{2}}}{\sqrt{x}} = \frac{x^{\frac{5}{2}}}{x^{\frac{1}{2}}}
$
Apply exponent rule (\( a^m / a^n = a^{m-n} \)):
$
x^{\frac{5}{2} – \frac{1}{2}} = x^{\frac{4}{2}} = x^2
$

Question Hard

4. Which expression is equivalent to \( b^{\frac{y}{x}} \), where \( b > 0 \)?

A) \( \sqrt[5]{b^{105}} \)
B) \( \sqrt[5]{b^{135}} \)
C) \( \sqrt[105]{b^{135}} \)
D) \( \sqrt[135]{b^{105}} \)

▶️Answer/Explanation

Answer: C
Convert to Radical Form:
$
b^{\frac{y}{x}} = \sqrt[x]{b^y}
$
The correct form is \( \sqrt[105]{b^{135}} \), since \( \frac{135}{105} = \frac{y}{x} \).

Question Hard

5. Which expression represents the product of \( (b^6 c^{-2} d^{-5}) \) and \( (b^8 c^{-3} + c^4 d^5) \), where \( b, c, \) and \( d \) are positive?

A) \( b^{14} c^{-5} + c^2 d^{-10} \)
B) \( b^{14} c^{-5} + c^2 \)
C) \( b^{14} c^{-5} d^{-5} + b^6 c^2 \)
D) \( b^{14} c^{-5} d^{-5} + c^2 \)

▶️Answer/Explanation

Answer: C
Multiply the Terms:
$
(b^6 c^{-2} d^{-5}) \cdot (b^8 c^{-3} + c^4 d^5)
$
Distribute:
$
b^6 c^{-2} d^{-5} \cdot b^8 c^{-3} + b^6 c^{-2} d^{-5} \cdot c^4 d^5
$
Simplify Each Term:
1. First term:
$
b^{6+8} c^{-2-3} d^{-5} = b^{14} c^{-5} d^{-5}
$
2. Second term:
$
b^6 c^{-2+4} d^{-5+5} = b^6 c^2 d^0 = b^6 c^2
$

Question Hard

6. If \( x \neq 0 \) and \( y \neq 0 \), which of the following is equivalent to \( \frac{8x^2}{\sqrt{4x^6 y^4}} \)?

A) \( 2xy^{-2} \)
B) \( 4x^{-2} y^2 \)
C) \( 4x^{-1} y^{-2} \)
D) \( 4xy^2 \)

▶️Answer/Explanation

Answer: 

Question Hard

7. If \( r \) and \( s \) are positive, which of the following expressions is equivalent to \( \frac{r^3}{s^7} \)?

A) \( \frac{1}{\sqrt[7]{r^5 s^{14}}} \)
B) \( \sqrt[7]{r^5 s^{14}} \)
C) \( \frac{1}{\sqrt[7]{r^6 s^3}} \)
D) \( \sqrt[7]{r^9 s^3} \)

▶️Answer/Explanation

Answer: 

Question Hard

8. The given equation \( \sqrt[d]{a} = \sqrt[b]{c} \) relates the distinct positive numbers \( a, b, c, \) and \( d \). Which equation correctly expresses \( c \) in terms of \( a, b, \) and \( d \)?

A) \( c = \frac{d}{a^{\frac{1}{b}}} \)
B) \( c = a^{d – h} \)
C) \( c = a^{\frac{d}{b}} \)
D) \( c = a^{\frac{b}{d}} \)

▶️Answer/Explanation

Answer: 

Question Hard

9. Which expression is equivalent to \( \sqrt{16a^3} \), where \( a > 0 \)?
A) \( 4a^{\frac{3}{2}} \)
B) \( 4a^{\frac{5}{3}} \)
C) \( 8a^{\frac{3}{4}} \)
D) \( 8a^{\frac{4}{3}} \)

▶️Answer/Explanation

Answer: 

Question Hard

10. Which expression is equivalent to \( \sqrt{r s} (\sqrt{r} + \sqrt{s}) \), where \( r \geq 0 \) and \( s \geq 0 \)?

A) \( \sqrt{r^2 s} + r s^{\frac{5}{2}} \)
B) \( r \sqrt{s} + s \sqrt{r} \)
C) \( rs \sqrt{r} + s \)
D) \( \sqrt{rs} + r + s \)

▶️Answer/Explanation

Answer: 

Question Hard

11. The expression \( \left( \sqrt{x^2} \right)^n \), where \( n \) is a constant, is equivalent to \( x^8 \). What is the value of \( n \)?

▶️Answer/Explanation

Answer: 

Question Hard

12. Which expression is equivalent to \( 2^{-2k} \cdot 3^k \)?

A) \( (3\sqrt{2})^k \)
B) \( \left( \frac{1}{36} \right)^k \)
C) \( \left( \frac{3}{4} \right)^k \)
D) \( \left( \frac{9}{4} \right)^k \)

▶️Answer/Explanation

Answer: 

Question Hard

13. Which expression is equivalent to \( a^{\frac{2}{6}} \cdot (a^{\frac{5}{3}})^{\frac{1}{2}} \), where \( a \) is positive?
A) \( \sqrt[3]{a^2} \)
B) \( \sqrt[3]{a^{16}} \)
C) \( \sqrt[3]{a^8} \)
D) \( \sqrt[3]{a^{14}} \)

▶️Answer/Explanation

Answer: 

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