【数学I】例題3.3.2：2次方程式の解2（One More）★★

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% 例題I3.3.2：2次方程式の解2（One More）★★
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次の2次方程式を解け．

(1) $-0.5x^2+3x-1.5=0$

(2) $\sqrt{2}x^2-8x+8\sqrt{2}=0$

(3) $(x-2)^2+4(x-2)-12=0$

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% 例題I3.3.2の解答（One More）★★
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(1) 両辺に$-2$を掛けると，$x^2-6x+3=0$

よって，

\[
x=\frac{-(-3)\pm\sqrt{(-3)^2-1\cdot 3}}{1}=3\pm\sqrt{6}
\]

(2) 両辺に$\sqrt{2}$を掛けると，$2x^2-8\sqrt{2}x+16=0$

したがって，$x^2-4\sqrt{2}x+8=0$

左辺を因数分解すると，$(x-2\sqrt{2})^2=0$

よって，$x=2\sqrt{2}$

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% 例題I3.3.2の別解
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両辺に$\sqrt{2}$を掛けると，$2x^2-8\sqrt{2}x+16=0$

したがって，$x^2-4\sqrt{2}x+8=0$

よって，

\[
x=\frac{-(-2\sqrt{2})\pm\sqrt{(-2\sqrt{2})^2-1\cdot 8}}{1}=2\sqrt{2}
\]

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% 例題I3.3.2の解答（One More）★★
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(3) $X=x-2$とおくと，$X^2+4X-12=0$

したがって，$(X+6)(X-2)=0$

ゆえに，$X=-6,2$

すなわち，$x-2=-6,2$

よって，$x=-4,4$

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% 例題I3.3.2の別解
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$(x-2)^2+4(x-2)-12=0$より，$\{(x-2)+6\}\{(x-2)-2\}=0$

したがって，$(x+4)(x-4)=0$

よって，$x=-4,4$

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% 問題I3.3.2：2次方程式の解2（One More）★★
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次の2次方程式を解け．

(1) $-0.25x^2+2x-1=0$

(2) $\sqrt{3}x^2-12x+12\sqrt{3}=0$

(3) $(x+1)^2-6(x+1)+5=0$

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% 問題I3.3.2の解答（One More）★★
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(1) 両辺に$-4$を掛けると，$x^2-8x+4=0$

よって，

\[
x=\frac{-(-4)\pm\sqrt{(-4)^2-1\cdot 4}}{1}=4\pm 2\sqrt{3}
\]

(2) 両辺に$\sqrt{3}$を掛けると，$3x^2-12\sqrt{3}x+36=0$

したがって，$x^2-4\sqrt{3}x+12=0$

左辺を因数分解すると，$(x-2\sqrt{3})^2=0$

よって，$x=2\sqrt{3}$

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% 問題I3.3.2の別解
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両辺に$\sqrt{3}$を掛けると，$3x^2-12\sqrt{3}x+36=0$

したがって，$x^2-4\sqrt{3}x+12=0$

よって，

\[
x=\frac{-(-2\sqrt{3})\pm\sqrt{(-2\sqrt{3})^2-1\cdot 12}}{1}=2\sqrt{3}
\]

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% 問題I3.3.2の解答（One More）★★
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(3) $X=x+1$とおくと，$X^2-6X+5=0$

したがって，$(X-5)(X-1)=0$

ゆえに，$X=5,1$

すなわち，$x+1=5,1$

よって，$x=4,0$

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% 問題I3.3.2の別解
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$(x+1)^2-6(x+1)+5=0$より，$\{(x+1)-5\}\{(x+1)-1\}=0$

したがって，$(x-4)\cdot x=0$

よって，$x=4,0$

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% 例題I3.3.2の側注：2次方程式の解2（One More）★★
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% 側注1
$x^2$の係数が正となるように，$-2$を掛ける．

% 側注2
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注3
$x^2$の係数が有理数となるように，$\sqrt{2}$を掛ける．

% 側注4
$(2\sqrt{2})^2=8$

% 側注5
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注6
共通する部分を1つのまとまりと見る．

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% 問題I3.3.2の側注：2次方程式の解2（One More）★★
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% 側注1
$x^2$の係数が正となるように，$-4$を掛ける．

% 側注2
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注3
$x^2$の係数が有理数となるように，$\sqrt{3}$を掛ける．

% 側注4
$(2\sqrt{3})^2=12$

% 側注5
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注6
共通する部分を1つのまとまりと見る．

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% 例題I3.3.2の考え方・補足など：2次方程式の解2（One More）★★
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% 考え方
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(1)，(2)は係数に小数や無理数が含まれており，そのまま解の公式を使うと計算に手間が掛かる．そこで，係数（とくに$x^2$の係数）は，正の数や有理数といった簡単な形となるように式変形するとよい．