Chapter 6.4, Problem 37WE

(a)

To determine

To Find:

The conjugate of $2\sqrt{5}-3\sqrt{2}$ .

Answer to Problem 37WE

  $2\sqrt{5}+3\sqrt{2}$

Explanation of Solution

Given:

Given expression is:

  $2\sqrt{5}-3\sqrt{2}$

Calculation:

The expressions of the form $a\sqrt{b}+c\sqrt{d}\text{ and }a\sqrt{b}-c\sqrt{d}$ are called conjugates .

So, the conjugate of $2\sqrt{5}-3\sqrt{2}\text{ is }2\sqrt{5}+3\sqrt{2}$ .

(b)

To determine

To Find:

The reciprocal of the conjugate of $2\sqrt{5}-3\sqrt{2}$ .

Answer to Problem 37WE

  $\frac{2\sqrt{5}-3\sqrt{2}}{2}$

Explanation of Solution

Given:

From part (a), the conjugate of $2\sqrt{5}-3\sqrt{2}\text{ is }2\sqrt{5}+3\sqrt{2}$ .

Formula Used:

  $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Calculation:

The conjugate of $2\sqrt{5}-3\sqrt{2}\text{ is }2\sqrt{5}+3\sqrt{2}$ .

Reciprocal of $x\text{ is }\frac{1}{x}$ .

So,

The reciprocal of $2\sqrt{5}+3\sqrt{2}\text{ is }\frac{1}{2\sqrt{5}+3\sqrt{2}}$ .

Since the denominator cannot have radicals, we need to rationalize the denominator by multiplying and dividing the numerator and denominator by the conjugate of the denominator.

  $\begin{array}{l}\frac{1}{2\sqrt{5}+3\sqrt{2}}\\ =\frac{1}{2\sqrt{5}+3\sqrt{2}}\cdot \frac{2\sqrt{5}-3\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}\\ =\frac{2\sqrt{5}-3\sqrt{2}}{\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\\ =\frac{2\sqrt{5}-3\sqrt{2}}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{2}\right)}^2}\mathrm{.........}[\left(a+b\right)\left(a-b\right)=a^2-b^2]\\ =\frac{2\sqrt{5}-3\sqrt{2}}{20-18}\\ =\frac{2\sqrt{5}-3\sqrt{2}}{2}\end{array}$

So, the reciprocal of the conjugate of $2\sqrt{5}-3\sqrt{2}\text{ is }\frac{2\sqrt{5}-3\sqrt{2}}{2}$ .

(c)

To determine

To Find:

The conjugate of the reciprocal of $2\sqrt{5}-3\sqrt{2}$ .

Answer to Problem 37WE

  $\sqrt{5}-\frac{3}{2}\sqrt{2}$

Explanation of Solution

Given:

Given expression:

  $2\sqrt{5}-3\sqrt{2}$

Formula Used:

  $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Calculation:

Reciprocal of $x\text{ is }\frac{1}{x}$ .

So,

The reciprocal of $2\sqrt{5}-3\sqrt{2}\text{ is }\frac{1}{2\sqrt{5}-3\sqrt{2}}$ .

Since the denominator cannot have radicals, we need to rationalize the denominator by multiplying and dividing the numerator and denominator by the conjugate of the denominator.

  $\begin{array}{l}\frac{1}{2\sqrt{5}-3\sqrt{2}}\\ =\frac{1}{2\sqrt{5}-3\sqrt{2}}\cdot \frac{2\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+3\sqrt{2}}\\ =\frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{\left(2\sqrt{5}-3\sqrt{2}\right)\left(2\sqrt{5}+3\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{2}\right)}^2}\mathrm{.................}[\left(a+b\right)\left(a-b\right)=a^2-b^2]\\ =\frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{20-18}\\ =\frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{2}\\ =\sqrt{5}+\frac{3}{2}\sqrt{2}\end{array}$

The expressions $a\sqrt{b}+c\sqrt{d}\text{ and }a\sqrt{b}-c\sqrt{d}$ are conjugates.

So, the conjugate of $\sqrt{5}+\frac{3}{2}\sqrt{2}\text{ is }\sqrt{5}-\frac{3}{2}\sqrt{2}$ .

Hence, the conjugate of the reciprocal of $2\sqrt{5}-3\sqrt{2}\text{ is }\sqrt{5}-\frac{3}{2}\sqrt{2}$ .