# 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

[No- Calc]  Question   Hard

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

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

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:
$x^2 – 14x + 49 + 2x – 14 = 0$
$x^2 – 12x + 35 = 0$

$(x – 5)(x – 7) = 0$

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 $$7$$.

[Calc]  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) $$\sqrt[5]{4x-4xy+y}$$

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:
$\boxed{(4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}}$

[No calc]  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. $$8\left(\sqrt[10]{2^7}\right)$$

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 $$2^{\frac{37}{10}}$$.

[Calc]  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) $\sqrt[3]{2 q+4 \sqrt{q r}+2 r}$

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

D

Question

Which of the following is equivalent to $\sqrt{16a^{16}}$

B

Questions

$x-2 \sqrt{x}-3=0$

What value of $x$ satisfies the equation above?

Ans: 9

Questions

$\sqrt{x+4}=11$

What value of $x$ satisfies the equation above?

Ans: 117

Questions

$K=\frac{200 v^2}{2}$

In the equation above, the kinetic energy, $K$, of a 200-gram object is given in terms of its speed, $v$. If the equation is rewritten in the form $v=a \sqrt{K}$, where $a$ is a positive constant, what is the value of $a$ ?

Ans: . $1,1 / 10$

Questions

If $\frac{\sqrt{x^5}}{\sqrt[3]{x^4}}=x^{\frac{a}{b}}$ for all positive values of $x$, what is the value of $\frac{a}{b}$ ? 2.7

Ans: $7 / 6,1.16,1.17$
If $r>0$ and $\sqrt[3]{\frac{9 r}{2}}=\frac{1}{2} r$, what is the value of $r$ ?