Chapter 1.5, Problem 34E

Solution

a.

To solve: the given equation or the inequality.

  $x\ge \frac{\sqrt{6}}{2}$ .

Given information:

The inequality given is $\sqrt{12}-x\sqrt{2}\le \sqrt{3}$ .

Property Used:

Dividing of multiplying an inequality by a negative number reverses the inequality.

  $\begin{array}{c}\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}\\ \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\end{array}$

Calculation:

Consider, $\sqrt{12}-x\sqrt{2}\le \sqrt{3}$ .

Subtract $\sqrt{12}$ from both sides, $-x\sqrt{2}\le \sqrt{3}-\sqrt{12}$ .

Divide both sides by $-\sqrt{2}$ (since it is a negative number, it will reverse the inequality)

  $\frac{-x\sqrt{2}}{-\sqrt{2}}\ge \frac{\sqrt{3}-\sqrt{12}}{-\sqrt{2}}$ .

This is equivalent to, $x\ge \frac{\sqrt{3}-\sqrt{12}}{-\sqrt{2}}$ .

Using the property of radicals that $\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}$ , the above inequality is equivalent to,

  $\begin{array}{c}x\ge \frac{\sqrt{3}-\sqrt{4\cdot 3}}{-\sqrt{2}}\\ x\ge \frac{\sqrt{3}-\sqrt{4}\cdot \sqrt{3}}{-\sqrt{2}}\\ x\ge \frac{\sqrt{3}-2\sqrt{3}}{-\sqrt{2}}\\ x\ge \frac{-\sqrt{3}}{-\sqrt{2}}\\ x\ge \frac{\sqrt{3}}{\sqrt{2}}\end{array}$

Now, rationalize the denominator,

  $\begin{array}{c}x\ge \frac{\sqrt{3}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\\ x\ge \frac{\sqrt{3\cdot 2}}{\sqrt{2\cdot 2}}\\ x\ge \frac{\sqrt{6}}{\sqrt{4}}\\ x\ge \frac{\sqrt{6}}{2}\end{array}$

The required solution is $x\ge \frac{\sqrt{6}}{2}$ .

b.

To solve: the given equation or the inequality.

  $x>\sqrt{3}+1$ .

Given information:

The inequality given is $x\sqrt{3}>2+x$ .

Property Used:

  $\begin{array}{c}\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}\\ \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\end{array}$

Calculation:

Consider, $x\sqrt{3}>2+x$ .

Subtract $x$ from both sides, $x\sqrt{3}-x>2$ .

This is equivalent to, $x\left(\sqrt{3}-1\right)>2$ .

Divide both sides by $\sqrt{3}-1$ .

  $\begin{array}{c}\frac{x\left(\sqrt{3}-1\right)}{\sqrt{3}-1}>\frac{2}{\sqrt{3}-1}\\ x>\frac{2}{\sqrt{3}-1}\end{array}$

Now, rationalize the denominator by multiplying by the conjugate of the denominator.

  $\begin{array}{c}x>\frac{2}{\sqrt{3}-1}\cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}\\ x>\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ x>\frac{2\left(\sqrt{3}+1\right)}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ x>\frac{2\left(\sqrt{3}+1\right)}{3-1}\\ x>\frac{2\left(\sqrt{3}+1\right)}{2}\\ x>\sqrt{3}+1\end{array}$

The required solution is $x>\sqrt{3}+1$ .

c.

To solve: the given equation or the inequality.

  $x=\frac{\sqrt{10}-2\sqrt{2}}{2}$ .

Given information:

The equality given is $2\sqrt{5}=4+x\sqrt{8}$ .

Property Used:

  $\begin{array}{c}\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}\\ \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\end{array}$

Calculation:

Consider, $2\sqrt{5}=4+x\sqrt{8}$ .

Subtract 4 from both sides, $2\sqrt{5}-4=x\sqrt{8}$ .

Divide both sides by $\sqrt{8}$ , $\frac{2\sqrt{5}-4}{\sqrt{8}}=\frac{x\sqrt{8}}{\sqrt{8}}$ .

This is equivalent to, $\frac{2\sqrt{5}-4}{\sqrt{8}}=x$ .

Now, rationalize the denominator by making it a perfect square.

  $\begin{array}{l}x=\frac{2\sqrt{5}-4}{\sqrt{8}}\\ x=\frac{2\sqrt{5}-4}{\sqrt{8}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\\ x=\frac{\left(2\sqrt{5}-4\right)\sqrt{2}}{\sqrt{16}}\\ x=\frac{\left(2\sqrt{5}\right)\sqrt{2}-4\sqrt{2}}{4}\\ x=\frac{2\sqrt{10}-4\sqrt{2}}{4}\\ x=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{4}\\ x=\frac{\sqrt{10}-2\sqrt{2}}{2}\end{array}$

The required solution is $x=\frac{\sqrt{10}-2\sqrt{2}}{2}$ .