【数学I】例題1.2.4：2重根号（One More）★★

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% 例題I1.2.4：2重根号（One More）★★
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次の2重根号を簡単な形にせよ．

(1) $\sqrt{9+2\sqrt{14}}$

(2) $\sqrt{7-2\sqrt{10}}$

(3) $\sqrt{6+4\sqrt{2}}$

(4) $\sqrt{8-\sqrt{48}}$

(5) $\sqrt{4+\sqrt{15}}$

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% 例題I1.2.4の解答（One More）★★
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(1) $\sqrt{9+2\sqrt{14}}=\sqrt{(7+2)+2\sqrt{7\times 2}}=\sqrt{7}+\sqrt{2}$

(2) $\sqrt{7-2\sqrt{10}}=\sqrt{(5+2)-2\sqrt{5\times 2}}=\sqrt{5}-\sqrt{2}$

(3) $\sqrt{6+4\sqrt{2}}=\sqrt{6+2\sqrt{8}}=\sqrt{(4+2)+2\sqrt{4\times 2}}=2+\sqrt{2}$

(4) $\sqrt{8-\sqrt{48}}=\sqrt{8-2\sqrt{12}}=\sqrt{(6+2)-2\sqrt{6\times 2}}=\sqrt{6}-\sqrt{2}$

(5)

\begin{align*}
\sqrt{4+\sqrt{15}}&=\sqrt{\frac{8+2\sqrt{15}}{2}}=\frac{\sqrt{(5+3)+2\sqrt{5\times 3}}}{\sqrt{2}}=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\
&=\frac{\sqrt{2}(\sqrt{5}+\sqrt{3})}{2}=\frac{\sqrt{10}+\sqrt{6}}{2}
\end{align*}

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% 問題I1.2.4：2重根号（One More）★★
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次の2重根号を簡単な形にせよ．

(1) $\sqrt{7-2\sqrt{12}}$

(2) $\sqrt{12+6\sqrt{3}}$

(3) $\sqrt{10-\sqrt{84}}$

(4) $\sqrt{8+3\sqrt{7}}$

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% 問題I1.2.4の解答（One More）★★
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(1) $\sqrt{7-2\sqrt{12}}=\sqrt{(4+3)-2\sqrt{4\times 3}}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}$

(2) $\sqrt{12+6\sqrt{3}}=\sqrt{(9+3)+2\sqrt{9\times 3}}=3+\sqrt{3}$

(3) $\sqrt{10-\sqrt{84}}=\sqrt{10-2\sqrt{21}}=\sqrt{(7+3)-2\sqrt{7\times 3}}=\sqrt{7}-\sqrt{3}$

(4)

\begin{align*}
\sqrt{8+3\sqrt{7}}&=\sqrt{\frac{16+2\sqrt{63}}{2}}=\frac{\sqrt{(9+7)+2\sqrt{9\times 7}}}{\sqrt{2}}\\
&=\frac{\sqrt{9}+\sqrt{7}}{\sqrt{2}}=\frac{3\sqrt{2}+\sqrt{14}}{2}
\end{align*}

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% 例題I1.2.4の側注：2重根号（One More）★★
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% 側注1
和が$9$，積が$14$になる2つの数は，$7$と$2$である．

% 側注2
$\sqrt{2}-\sqrt{5}$は誤りであるので注意すること．

% 側注3
\begin{align*}
4\sqrt{2}&=2\times 2\sqrt{2}\\
&=2\sqrt{2^2\cdot 2}=2\sqrt{8}
\end{align*}

% 側注4
$\sqrt{48}=\sqrt{2^2\cdot 12}=2\sqrt{12}$

% 側注5
$\frac{\sqrt{4+\sqrt{15}}}{1}$の分母，分子に$\sqrt{2}$を掛ける．

% 側注6
分母を有理化する．

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% 問題I1.2.4の側注：2重根号（One More）★★
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% 側注1
$\sqrt{3}-2$は誤りであるので注意すること．

% 側注2
\begin{align*}
6\sqrt{3}&=2\times 3\sqrt{3}\\
&=3\sqrt{3^2\cdot 3}=3\sqrt{27}
\end{align*}

% 側注3
$\sqrt{84}=\sqrt{2^2\cdot 21}=2\sqrt{21}$

% 側注4
$\frac{\sqrt{8+3\sqrt{7}}}{1}$の分母，分子に$\sqrt{2}$を掛ける．

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% 例題I1.2.4の考え方・補足など：2重根号（One More）★★
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% 考え方
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$\sqrt{p\pm 2\sqrt{q}}$の形は，$a+b=p,ab=q$（和が$p$，積が$q$）になる数$a,b$を見つければ，次のように式変形できる．$a>0,b>0$のとき，$\sqrt{p+2\sqrt{q}}=\sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b},$

$a>b>0$のとき，$\sqrt{p-2\sqrt{q}}=\sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}$

(1) 和が$9$，積が$14$になる2つの数を探す．

(2) 和が$7$，積が$10$になる2つの数を探す．

(3)，(4)，(5)$2\sqrt{ab}$の形を作るように式変形する．

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\begin{align*}
&\sqrt{p+2\sqrt{q}}=\sqrt{a+b+2\sqrt{ab}}=\sqrt{(\sqrt{a}+\sqrt{b})^2}=\sqrt{a}+\sqrt{b}(a>0,b>0),\\
&\sqrt{p-2\sqrt{q}}=\sqrt{a+b-2\sqrt{ab}}=\sqrt{(\sqrt{a}-\sqrt{b})^2}=\sqrt{a}-\sqrt{b}(a>b>0)
\end{align*}

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% 注意
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$2\sqrt{ab}$の$2$がない場合は，$2$を作るように式変形する．

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% 注意
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$\sqrt{(a+b)-2\sqrt{ab}}=\sqrt{(\sqrt{a}-\sqrt{b})^2}=\sqrt{a}-\sqrt{b}$のときの2つの数$a,b$は，$a>b>0$であるので注意すること．例えば(2)は，$\sqrt{2}>\sqrt{5}$ではなく，$\sqrt{5}>\sqrt{2}$であるので，$\sqrt{7-2\sqrt{10}}=\sqrt{2}-\sqrt{5}$と答えてしまわないようにする（正しくは，$\sqrt{7-2\sqrt{10}}=\sqrt{5}-\sqrt{2}$）．