Question

find the value of
√(6-2√5)

Answer 1

Answer:

$\sqrt{5}-1$

Step-by-step explanation:

Step 1: $\sqrt{6-2 \sqrt{5}}$

Apply rule: (a)=a

$\begin{aligned}& (6-2 \sqrt{5})=6-2 \sqrt{5} \\& =\sqrt{6-2 \sqrt{5}} \\& =\sqrt{5-2 \sqrt{5}+1} \\& =\sqrt{1 \cdot 5-2 \sqrt{5}+1} \\& =\sqrt{(\sqrt{1})^2(\sqrt{5})^2-2 \sqrt{5}+(\sqrt{1})^2}\end{aligned}$

$\begin{aligned}& \sqrt{1}=1 \\& =\sqrt{1^2(\sqrt{5})^2-2 \sqrt{5}+1^2}\end{aligned}$

$\begin{aligned}& 2 \cdot 1 \cdot 1 \cdot \sqrt{5}=2 \sqrt{5} \\& =\sqrt{1^2(\sqrt{5})^2-2 \cdot 1 \cdot 1 \cdot \sqrt{5}+1^2}\end{aligned}$

Step 2: To remove quantities from radical expressions, apply the condition n√xn=x x n n = x if n is odd and n√xn=|x| x n n = | x | if n is even. To simplify radical equations, seek for exponential factors within the radical. When simplifying radical formulas, all exponentiation and integer operation rules apply.

Apply Perfect Square Formula:$(a-b)^2=a^2-2 a b+b^2$

$1^2(\sqrt{5})^2-2 \cdot 1 \cdot 1 \cdot \sqrt{5}+1^2=(1 \cdot \sqrt{5}-1)^2$

$=\sqrt{(1 \cdot \sqrt{5}-1)^2}$

Apply radical rule: $\quad \sqrt[n]{a^n}=a$

$\begin{aligned}& \sqrt{(1 \cdot \sqrt{5}-1)^2}=1 \cdot \sqrt{5}-1 \\& =1 \cdot \sqrt{5}-1\end{aligned}$

Apply rule: $1 \cdot a=a$

$\begin{aligned}& 1 \cdot \sqrt{5}=\sqrt{5} \\& =\sqrt{5}-1\end{aligned}$

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