Chapter 6.4, Problem 25WE

To determine

To Simplify:

  ${\left(\sqrt{6}-\sqrt{15}\right)}^2$

Answer to Problem 25WE

  $21-6\sqrt{10}$

Explanation of Solution

Given:

The expression is :

  ${\left(\sqrt{6}-\sqrt{15}\right)}^2$

Formula Used:

  ${\left(a-b\right)}^2=a^2-2ab+b^2$

Calculation:

Given:

  ${\left(\sqrt{6}-\sqrt{15}\right)}^2\mathrm{..............}(1)$

Use

  ${\left(a-b\right)}^2=a^2-2ab+b^2\mathrm{..........}(2)$

Comparing equation (1) with (2),

  $a=\sqrt{6},b=\sqrt{15}$

Substitute the values in equation (2),

  $\begin{array}{l}{\left(a-b\right)}^2\\ =a^2-2ab+b^2\\ \Rightarrow {\left(\sqrt{6}-\sqrt{15}\right)}^2\\ ={\left(\sqrt{6}\right)}^2-2\left(\sqrt{6}\right)\left(\sqrt{15}\right)+{\left(\sqrt{15}\right)}^2\\ =6-2\sqrt{2\cdot 3\cdot 5\cdot 3}+\mathrm{15.................}[\text{Product property of radicals;}\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}]\\ =21-2\cdot 3\cdot \sqrt{10}\\ =21-6\sqrt{10}\end{array}$