## Math 9 Chapter 1 Lesson 4: Relationship between division and squares

## 1. Summary of theory

### 1.1. Theorem

For non-negative a and positive b, we have: \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)

### 1.2. Apply

**a. The rule of squaring a quotient**

To square a quotient \(\frac{a}{b}\), where a is not negative and b is positive, we can square the roots of a and b, respectively, and divide the first result. for the second result.

**b. Rule for dividing two square roots**

To divide the square root of a non-negative number a and a positive number b, we can divide the number a by the number b and square the result we just found.

## 2. Illustrated exercise

### 2.1. Type 1: Problem using square one quotient rule

**Question 1:** Simplify the following expression:

\(5xy.\sqrt{\frac{25x^2}{y^6}}\) with \(x> 0; y\neq 0\) ; \(0.2x^3y^3\sqrt{\frac{16}{x^4y^8}}\) with \(x\neq 0;y\neq 0\)

**Solution guide**

** **\(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\frac{5|x|}{y^3}=\frac{25x^2y}{y^3}= \frac{25x^2}{y^2}\)

Similarly, we have: \(0.2x^3y^3\sqrt{\frac{16}{x^4y^8}}=\frac{0.2x^3y^3.4}{x^2y^4} =\frac{0.8x}{y}\)

**Verse 2: **Simplify the following expression:

\(\sqrt{\frac{27(a-3)^2}{48}}\) with \(a>3\) ; \((ab).\sqrt{\frac{ab}{(ab)^2}}\) with (a < b)

**Solution guide**

\(\sqrt{\frac{27(a-3)^2}{48}}=\sqrt{\frac{9}{16}}|a-3|=\frac{3}{4}(a -3)\) (because \(a>3\) so \(a-3>0\))

\((ab).\sqrt{\frac{ab}{(ab)^2}}=(ab)\frac{\sqrt{ab}}{|ab|}=(ab)\frac{\sqrt{ ab}}{ba}=-\sqrt{ab}\) (because a < b)

### 2.2. Type 2: Problem using the rule of dividing two square roots

**Question 1: **Perform the following calculations: \(\frac{\sqrt{52}}{\sqrt{117}}\) ; \(\frac{\sqrt{2}}{\sqrt{18}}\)

**Solution guide**

We have: \(\frac{\sqrt{52}}{\sqrt{117}}=\sqrt{\frac{52}{117}}=\sqrt{\frac{4}{9}}=\frac {2}{3}\)

Similarly, we have \(\frac{\sqrt{2}}{\sqrt{18}}=\sqrt{\frac{2}{18}}=\sqrt{\frac{1}{9}}= \frac{1}{3}\)

**Verse 2:** Solve the equation: \(\sqrt{2}x-\sqrt{50}=0\) ; \(\frac{x^2}{5}-\sqrt{20}=0\)

**Solution guide**

\(\sqrt{2}x-\sqrt{50}=0\Leftrightarrow \sqrt{2}x=\sqrt{50}\Leftrightarrow x=\frac{\sqrt{50}}{\sqrt{2}} =5\)

Similarly, we have: \(\frac{x^2}{5}-\sqrt{20}=0\Leftrightarrow \frac{x^2}{5}=\sqrt{20}\Leftrightarrow x^2= 5\sqrt{20}\Leftrightarrow x=\pm \sqrt{\sqrt{500}}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Perform the following calculations: \(\frac{{\sqrt {72} }}{{\sqrt {120} }}\) ; \(\frac{{\sqrt 5 }}{{\sqrt {75} }}\)

**Verse 2: **Simplify the following expression: \(3xy.\sqrt {\frac{{9{y^2}}}{{{x^6}}}} \) with \(x \ne 0;\,y > 0\) ; \(0.4{x^3}{y^3}\sqrt {\frac{{25}}{{{x^8}{y^4}}}} \) with \(x \ne 0;y \ne 0\)

**Question 3: **Solve the equation: \(\sqrt 3 x – \sqrt {108} = 0\) ; \(\frac{{{x^2}}}{{\sqrt 5 }} – \sqrt {20} = 0\)

**Question 4: **Simplify the following expression: \(\sqrt {\frac{{54{{(2 – x)}^2}}}{{24}}} \) with \(x > 2\) ; \({(a – b)^2}.\sqrt {\frac{{{a^4}{b^8}}}{{{{(a – b)}^6}}}} \) with a > b

**Question 5:** Solve the equation: \(\sqrt {{x^2} – 4x + 13} = 3\)

### 3.2. Multiple choice exercises

**Question 1.** Which of these following statements is wrong?

A. \(\frac{{\sqrt 3 }}{{\sqrt 7 }} = \frac{1}{3}\)

B. \(\frac{{\sqrt {15} }}{{\sqrt {735} }} = \frac{1}{7}\)

C. \(\frac{{\sqrt {480000} }}{{\sqrt {300} }} = 4\)

D. \(\frac{{\sqrt {{{12}^5}} }}{{\sqrt {{2^3}{6^5}} }} = 2\)

**Verse 2. **Calculate \(M = \sqrt {1,69.1,38 – 1,69.0,74} \)

A.1.04

B. 1.64

C. 2.08

D. 2.14

**Verse 3.** Calculate \(N = \sqrt {\frac{{{{125}^2} – {{100}^2}}}{{400}}} \)

A. 15/2

B. 1/15

C. 5/4

D. Other results

**Verse 4.** The solution of the equation \(\sqrt{5}x+\sqrt{5}=\sqrt{20}+\sqrt{45}\) is

A. 2 B. 3

C. 4 D. 5

**Question 5.** Without using a calculator, the value of the expression \(\sqrt{\frac{149^2-76^2}{457^2-384^2}}\) is

A. \(\frac{13}{29}\)

B. \(\frac{13}{27}\)

C. \(\frac{15}{27}\)

D. \(\frac{15}{29}\)

## 4. Conclusion

Through this lesson, you should know the following:

- Know the rules of squaring a quotient, the rules of dividing square roots.
- Do related maths.

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