Chapter 7.4, Problem 72PPS

To determine

To simplify the expression ${(\sqrt{2}+\sqrt{3})}^2$ .

Answer to Problem 72PPS

  $5-6\sqrt{6}$

Explanation of Solution

Given:

  ${(\sqrt{2}+\sqrt{3})}^2$

Formula used:

The radical rules:

  ${(a-b)}^2=a^2+b^2-2ab$

  $\sqrt[n]{a}=a^{\frac{1}{n}}$

  ${(a^m)}^n=a^{mn}$

  $\sqrt[a]{b\times c}=\sqrt[a]{b}\times \sqrt[a]{c}$

  $a^m\cdot a^n=a^{m+n}$

Calculation:

  $\begin{array}{l}$ \Rightarrow {(\sqrt{2}+\sqrt{3})}^2$$ \Rightarrow {(\sqrt{2})}^2+{(\sqrt{3})}^2-2\times \sqrt{2}\times \sqrt{3}[\because {(a-b)}^2=a^2+b^2-2ab]$$ \Rightarrow {(2^{\frac{1}{2}})}^2+{(3^{\frac{1}{2}})}^2-2\times \sqrt{2}\times \sqrt{3}[\because \sqrt[n]{a}=a^{\frac{1}{n}}]$$ \Rightarrow {(2)}^{\frac{2}{2}}+{(3)}^{\frac{2}{2}}-2\times \sqrt{2}\times \sqrt{3}[\because a^m\cdot a^n=a^{m+n}]$\Rightarrow 2+3-2\times \sqrt{6}\text{ [}\because \sqrt[a]{b\times c}=\sqrt[a]{b}\times \sqrt[a]{c}\text{] }\\ \Rightarrow 5-6\sqrt{6}\end{array}$

Conclusion:

The simplified form of the given expression is equal to $5-6\sqrt{6}.$