exercises
these will be similar to what you will be tested on in the placement test into grade 12 (math 30-1/math 182) math. note that the sequences & series unit is not on this guide and is not on the placement test.
all questions below should be completed without a calculator.
1) simplify each of the following expressions:
a) $\sqrt{\frac{2}{7}}$
b) $\sqrt{9^2 + 5^2}$
c) $\sqrt3{\frac{16a^4}{27}}$
d) $ 3cd\sqrt{12a^3b^6c^7} $
2) perform the following operations and simplify:
a) $\left(3\sqrt{2} - 4\sqrt{6}\
ight)^2$
b) $\frac{4 - \sqrt{10}}{\sqrt{72}}$
c) $\frac{\sqrt{3}}{3\sqrt{6} + \sqrt{2}}$
3) solve the following radical equations (do not include extraneous answers):
a) $\sqrt{x - 3} = 5$

Question

exercises
these will be similar to what you will be tested on in the placement test into grade 12 (math 30-1/math 182) math. note that the sequences & series unit is not on this guide and is not on the placement test.
all questions below should be completed without a calculator.
1) simplify each of the following expressions:
   a) $\sqrt{\frac{2}{7}}$
   b) $\sqrt{9^2 + 5^2}$
   c) $\sqrt3{\frac{16a^4}{27}}$
   d) $ 3cd\sqrt{12a^3b^6c^7} $
2) perform the following operations and simplify:
   a) $\left(3\sqrt{2} - 4\sqrt{6}\
ight)^2$
   b) $\frac{4 - \sqrt{10}}{\sqrt{72}}$
   c) $\frac{\sqrt{3}}{3\sqrt{6} + \sqrt{2}}$
3) solve the following radical equations (do not include extraneous answers):
   a) $\sqrt{x - 3} = 5$

Accepted Answer

### 1) a) Simplify $\boldsymbol{\sqrt{\frac{2}{7}}}$
# Explanation:
## Step1: Rationalize the denominator
To rationalize the denominator of a square root fraction, we multiply the numerator and denominator inside the square root by the denominator. So, $\sqrt{\frac{2}{7}}=\sqrt{\frac{2	imes7}{7	imes7}}$
## Step2: Simplify the square root
Simplify the square root. $\sqrt{\frac{14}{49}}=\frac{\sqrt{14}}{\sqrt{49}}=\frac{\sqrt{14}}{7}$
# Answer:
$\frac{\sqrt{14}}{7}$

### 1) b) Simplify $\boldsymbol{\sqrt{9^{2}+5^{2}}}$
# Explanation:
## Step1: Calculate the squares
First, calculate the squares of 9 and 5. $9^{2} = 81$ and $5^{2}=25$
## Step2: Add the results
Add the two results: $81 + 25=106$
## Step3: Take the square root
Take the square root of 106. $\sqrt{106}$ (since 106 has no perfect square factors other than 1)
# Answer:
$\sqrt{106}$

### 1) c) Simplify $\boldsymbol{\sqrt[3]{\frac{16a^{4}}{27}}}$
# Explanation:
## Step1: Split the cube root
We can split the cube root into the cube root of the numerator divided by the cube root of the denominator. $\sqrt[3]{\frac{16a^{4}}{27}}=\frac{\sqrt[3]{16a^{4}}}{\sqrt[3]{27}}$
## Step2: Simplify the denominator
The cube root of 27 is 3, since $3^{3}=27$. So, $\frac{\sqrt[3]{16a^{4}}}{3}$
## Step3: Simplify the numerator
We can rewrite $16a^{4}$ as $8	imes2	imes a^{3}	imes a$. Then $\sqrt[3]{8	imes2	imes a^{3}	imes a}=\sqrt[3]{8}	imes\sqrt[3]{a^{3}}	imes\sqrt[3]{2a}=2a\sqrt[3]{2a}$
## Step4: Combine the results
Substitute the simplified numerator back into the fraction: $\frac{2a\sqrt[3]{2a}}{3}$
# Answer:
$\frac{2a\sqrt[3]{2a}}{3}$

### 1) d) Simplify $\boldsymbol{3cd\sqrt{12a^{3}b^{6}c^{7}}}$
# Explanation:
## Step1: Simplify the square root
First, simplify the square root of $12a^{3}b^{6}c^{7}$. We can factor 12 as $4	imes3$, $a^{3}$ as $a^{2}	imes a$, $b^{6}$ as $(b^{3})^{2}$, and $c^{7}$ as $c^{6}	imes c$. So, $\sqrt{12a^{3}b^{6}c^{7}}=\sqrt{4	imes3	imes a^{2}	imes a	imes(b^{3})^{2}	imes c^{6}	imes c}$
## Step2: Take out the perfect squares
Take out the perfect squares from the square root: $2ab^{3}c^{3}\sqrt{3ac}$
## Step3: Multiply by the outside term
Now, multiply this by $3cd$: $3cd	imes2ab^{3}c^{3}\sqrt{3ac}$
## Step4: Simplify the product
Multiply the coefficients and the like - variables: $6ab^{3}c^{4}d\sqrt{3ac}$
# Answer:
$6ab^{3}c^{4}d\sqrt{3ac}$

### 2) a) Simplify $\boldsymbol{(3\sqrt{2}-4\sqrt{6})^{2}}$
# Explanation:
## Step1: Use the formula $(a - b)^{2}=a^{2}-2ab + b^{2}$
Let $a = 3\sqrt{2}$ and $b = 4\sqrt{6}$. Then, $(3\sqrt{2}-4\sqrt{6})^{2}=(3\sqrt{2})^{2}-2	imes(3\sqrt{2})	imes(4\sqrt{6})+(4\sqrt{6})^{2}$
## Step2: Calculate each term
- Calculate $(3\sqrt{2})^{2}$: $3^{2}	imes(\sqrt{2})^{2}=9	imes2 = 18$
- Calculate $2	imes(3\sqrt{2})	imes(4\sqrt{6})$: $2	imes3	imes4	imes\sqrt{2	imes6}=24\sqrt{12}$
- Calculate $(4\sqrt{6})^{2}$: $4^{2}	imes(\sqrt{6})^{2}=16	imes6 = 96$
## Step3: Simplify $\sqrt{12}$
Simplify $\sqrt{12}=\sqrt{4	imes3}=2\sqrt{3}$, so $24\sqrt{12}=24	imes2\sqrt{3}=48\sqrt{3}$
## Step4: Combine the terms
Now, combine the three terms: $18-48\sqrt{3}+96$
## Step5: Combine like - terms
Combine the constant terms: $114 - 48\sqrt{3}$
# Answer:
$114 - 48\sqrt{3}$

### 2) b) Simplify $\boldsymbol{\frac{4-\sqrt{10}}{\sqrt{72}}}$
# Explanation:
## Step1: Simplify the denominator
First, simplify $\sqrt{72}$. We can factor 72 as $36	imes2$, so $\sqrt{72}=\sqrt{36	imes2}=6\sqrt{2}$
## Step2: Rationalize the denominator
Multiply the numerator and denominator by $\sqrt{2}$ to rationalize the denominator: $\frac{(4 - \sqrt{10})	imes\sqrt{2}}{6\sqrt{2}	imes\sqrt{2}}$
## Step3: Simplify the denominator
The denominator becomes $6	imes2 = 12$
## Step4: Simplify the numerator
Expand the numerator: $4\sqrt{2}-\sqrt{20}$
## Step5: Simplify $\sqrt{20}$
Simplify $\sqrt{20}=\sqrt{4	imes5}=2\sqrt{5}$
## Step6: Combine the results
So, the expression becomes $\frac{4\sqrt{2}-2\sqrt{5}}{12}$
## Step7: Factor out the common factor
Factor out a 2 from the numerator: $\frac{2(2\sqrt{2}-\sqrt{5})}{12}$
## Step8: Simplify the fraction
Simplify the fraction by dividing numerator and denominator by 2: $\frac{2\sqrt{2}-\sqrt{5}}{6}$
# Answer:
$\frac{2\sqrt{2}-\sqrt{5}}{6}$

### 2) c) Simplify $\boldsymbol{\frac{\sqrt{3}}{3\sqrt{6}+\sqrt{2}}}$
# Explanation:
## Step1: Rationalize the denominator
Multiply the numerator and denominator by the conjugate of the denominator, which is $3\sqrt{6}-\sqrt{2}$. So, $\frac{\sqrt{3}(3\sqrt{6}-\sqrt{2})}{(3\sqrt{6}+\sqrt{2})(3\sqrt{6}-\sqrt{2})}$
## Step2: Expand the denominator
Use the difference of squares formula $(a + b)(a - b)=a^{2}-b^{2}$ for the denominator. Here, $a = 3\sqrt{6}$ and $b=\sqrt{2}$. So, $(3\sqrt{6})^{2}-(\sqrt{2})^{2}=9	imes6 - 2=54 - 2 = 52$
## Step3: Expand the numerator
Expand the numerator: $\sqrt{3}	imes3\sqrt{6}-\sqrt{3}	imes\sqrt{2}=3\sqrt{18}-\sqrt{6}$
## Step4: Simplify $\sqrt{18}$
Simplify $\sqrt{18}=\sqrt{9	imes2}=3\sqrt{2}$, so the numerator becomes $3	imes3\sqrt{2}-\sqrt{6}=9\sqrt{2}-\sqrt{6}$
## Step5: Combine the results
So, the expression is $\frac{9\sqrt{2}-\sqrt{6}}{52}$
# Answer:
$\frac{9\sqrt{2}-\sqrt{6}}{52}$

### 3) a) Solve $\boldsymbol{\sqrt{x - 3}=5}$
# Explanation:
## Step1: Eliminate the square root
To eliminate the square root, we square both sides of the equation. $(\sqrt{x - 3})^{2}=5^{2}$
## Step2: Simplify both sides
Simplify both sides. The left - hand side simplifies to $x - 3$ and the right - hand side simplifies to 25. So, $x-3 = 25$
## Step3: Solve for x
Add 3 to both sides of the equation: $x=25 + 3=28$
# Answer:
$x = 28$

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QUESTION IMAGE

Question

Explanation:

1) a) Simplify $\boldsymbol{\sqrt{\frac{2}{7}}}$

Step1: Rationalize the denominator

Step2: Simplify the square root

Step1: Calculate the squares

Step2: Add the results

Step3: Take the square root

Step1: Split the cube root

Step2: Simplify the denominator

Step3: Simplify the numerator

Step4: Combine the results

Answer:

1) b) Simplify $\boldsymbol{\sqrt{9^{2}+5^{2}}}$