🧮 Algebra Radicals And Conjugates (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Basic Radicals · Easy · source: Original (AMC-style)
What is $\sqrt{16}$?
A) $2$
B) $4$
C) $8$
D) $16$
E) $32$
Answer & Solution
Answer: B
$\sqrt{16} = 4$ since $4^2 = 16$.
2.
Tags: Basic Radicals · Easy · source: Original (AMC-style)
What is $\sqrt{25} + \sqrt{9}$?
A) $5$
B) $7$
C) $8$
D) $13$
E) $34$
Answer & Solution
Answer: C
$\sqrt{25} + \sqrt{9} = 5 + 3 = 8$.
3.
Tags: Radical Multiplication · Easy · source: Original (AMC-style)
What is $\sqrt{2} \cdot \sqrt{8}$?
A) $2$
B) $4$
C) $8$
D) $16$
E) $32$
Answer & Solution
Answer: B
$\sqrt{2} \cdot \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4$.
4.
Tags: Radical Division · Easy · source: Original (AMC-style)
What is $\frac{\sqrt{12}}{\sqrt{3}}$?
A) $2$
B) $3$
C) $4$
D) $6$
E) $9$
Answer & Solution
Answer: A
$\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2$.
5.
Tags: Simplifying Radicals · Medium · source: Original (AMC-style)
What is $\sqrt{18}$ simplified?
A) $2\sqrt{3}$
B) $3\sqrt{2}$
C) $6\sqrt{2}$
D) $9\sqrt{2}$
E) $18$
Answer & Solution
Answer: B
$\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$.
6.
Tags: Radical Addition · Medium · source: Original (AMC-style)
What is $\sqrt{8} + \sqrt{18}$?
A) $5\sqrt{2}$
B) $6\sqrt{2}$
C) $7\sqrt{2}$
D) $8\sqrt{2}$
E) $9\sqrt{2}$
Answer & Solution
Answer: A
$\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$.
7.
Tags: Rationalizing · Medium · source: Original (AMC-style)
What is $\frac{1}{\sqrt{2}}$ rationalized?
A) $\frac{\sqrt{2}}{2}$
B) $\frac{\sqrt{2}}{4}$
C) $\frac{2}{\sqrt{2}}$
D) $\frac{4}{\sqrt{2}}$
E) $\sqrt{2}$
Answer & Solution
Answer: A
$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
8.
Tags: Conjugate · Medium · source: Original (AMC-style)
What is the conjugate of $3 + \sqrt{5}$?
A) $3 - \sqrt{5}$
B) $-3 + \sqrt{5}$
C) $-3 - \sqrt{5}$
D) $3 + \sqrt{5}$
E) $\sqrt{5} - 3$
Answer & Solution
Answer: A
The conjugate of $a + b\sqrt{c}$ is $a - b\sqrt{c}$, so the conjugate of $3 + \sqrt{5}$ is $3 - \sqrt{5}$.
9.
Tags: Complex Radicals · Hard · source: Original (AMC-style)
What is $\sqrt{2 + \sqrt{3}} \cdot \sqrt{2 - \sqrt{3}}$?
A) $1$
B) $\sqrt{3}$
C) $2$
D) $\sqrt{5}$
E) $3$
Answer & Solution
Answer: A
Using the difference of squares: $\sqrt{2 + \sqrt{3}} \cdot \sqrt{2 - \sqrt{3}} = \sqrt{(2 + \sqrt{3})(2 - \sqrt{3})} = \sqrt{4 - 3} = \sqrt{1} = 1$.
10.
Tags: Rationalizing Complex · Hard · source: Original (AMC-style)
What is $\frac{1}{2 + \sqrt{3}}$ rationalized?
A) $2 - \sqrt{3}$
B) $\frac{2 - \sqrt{3}}{1}$
C) $\frac{2 - \sqrt{3}}{7}$
D) $\frac{2 + \sqrt{3}}{7}$
E) $\frac{1}{2 - \sqrt{3}}$
Answer & Solution
Answer: A
Multiplying by the conjugate: $\frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.
11.
Tags: Nested Radicals · Hard · source: Original (AMC-style)
What is $\sqrt{6 + 2\sqrt{5}}$?
A) $1 + \sqrt{5}$
B) $2 + \sqrt{5}$
C) $\sqrt{5} + 1$
D) $\sqrt{5} + 2$
E) $3 + \sqrt{5}$
Answer & Solution
Answer: A
We need to find $a$ and $b$ such that $(a + b\sqrt{5})^2 = 6 + 2\sqrt{5}$. Expanding: $a^2 + 5b^2 + 2ab\sqrt{5} = 6 + 2\sqrt{5}$. So $a^2 + 5b^2 = 6$ and $2ab = 2$, giving $ab = 1$. Trying $a = 1, b = 1$: $(1 + \sqrt{5})^2 = 1 + 5 + 2\sqrt{5} = 6 + 2\sqrt{5}$ ✓.
12.
Tags: Advanced Conjugates · Hard · source: Original (AMC-style)
If $x = 1 + \sqrt{2}$, what is $x + \frac{1}{x}$?
A) $2$
B) $2\sqrt{2}$
C) $3$
D) $4$
E) $2 + \sqrt{2}$
Answer & Solution
Answer: A
We have $x = 1 + \sqrt{2}$, so $\frac{1}{x} = \frac{1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = -1 + \sqrt{2}$. Therefore $x + \frac{1}{x} = (1 + \sqrt{2}) + (-1 + \sqrt{2}) = 2\sqrt{2}$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | B | C | B | A | B | A | A | A | A | A | A | B |