Question 1

$\sqrt{29-12√(5)$ simplify this

Question 2

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

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

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

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

$\large\displaystyle\text{$\begin{gathered}\sf √(4)=2 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf √(9)=3 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 2\cdot2\cdot3√(5)=12√(5) \end{gathered}$}$

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

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

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

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

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

$\sqrt{(2√(5)-3)^(2) } =2√(5 )-3.$

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

↓

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

Question 3

$\sqrt{29 \: - \: 12 √(5) }$

$\sqrt{(3 \: - \: 2 √(5)) ^(2) }$

$\boxed{ \bold{2 √(5) \: - \: 3}}$

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