Question

qrt{5}" align="absmiddle" class="latex-formula"> simplify this

Answer 1

$\sqrt{29 \: - \: 12 \sqrt{5} }$

• Factor the indicated expression:

$\sqrt{(3 \: - \: 2 \sqrt{5}) ^{2} }$

• Simplified the index the root and also the exponent using the number 2.

$\boxed{ \bold{2 \sqrt{5} \: - \: 3}}$

MissSpanish

Answer 2

$\large\displaystyle\text{ \begin{gathered}\sf \sqrt{29-12\sqrt{5}} \end{gathered} }$

                                                       

$\large\displaystyle\text{ \begin{gathered}\sf =\sqrt{20-12\sqrt{5}+9} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\sf =\sqrt{4\cdot \:5-12\sqrt{5}+9} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\sf =\sqrt{\left(\sqrt{4}\right)^2\left(\sqrt{5}\right)^2-12\sqrt{5}+\left(\sqrt{9}\right)^2} \end{gathered} }$

                                                                                     

$\large\displaystyle\begin{gathered}\sf \sqrt{4}=2 \end{gathered}$

$\large\displaystyle\begin{gathered}\sf \sqrt{9}=3 \end{gathered}$

$\large\displaystyle\begin{gathered}\sf 2\cdot2\cdot3\sqrt{5}=12\sqrt{5} \end{gathered}$

                                                                                     

$\large\displaystyle\text{ \begin{gathered}\sf =\sqrt{2^2\left(\sqrt{5}\right)^2-2\cdot \:2\cdot \:3\sqrt{5}+3^2} \end{gathered} }$

$\large\displaystyle\begin{gathered}\sf Apply\:the\:formula\:from\:binomial\:to\:square:(a-b)^2=a^2-2ab+b^2 \end{gathered}$

$\large\displaystyle\begin{gathered}\sf 2^{2}(\sqrt{5})^{2}-2\cdot2\cdot3\sqrt{5}+3^{2}=(2\sqrt{5}-3)^{2} \end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\sf =\sqrt{\left(2\sqrt{5}-3\right)^2} \end{gathered} }$

                                                                                         

$\large\displaystyle\text{ \begin{gathered}\sf Apply \ the \ laws \ of \ exponents:\sqrt[n]{a^{2} }=a \end{gathered} }$

$\sqrt{(2\sqrt{5}-3)^{2} } =2\sqrt{5 }-3.$

$\red{\boxed{\large\displaystyle\text{ \begin{gathered}\sf \red{=2\sqrt{5}-3 } \end{gathered} }}} \ \ \red{\Rightarrow} \ \ \bf{\red{Answer}}$

↓

$\red{\boxed{\boxed{\large\displaystyle\text{ \begin{gathered}\sf \bf{\red{MyHeritage}} \end{gathered} }}}}$