【数学I】例題3.3.8：2次方程式の共通解（One More）★★★

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% 例題I3.3.8：2次方程式の共通解（One More）★★★
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$x$についての2つの2次方程式$x^2+(k-6)x-4=0,x^2-2x-k=0$がただ1つの共通な実数解をもつとき，定数$k$の値と，そのときの共通解を求めよ．

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% 例題I3.3.8の解答（One More）★★★
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共通解を$\alpha$として，2つの2次方程式に$x=\alpha$を代入すると，

\[
\{\begin{array}{l}
\alpha^2+(k-6)\alpha-4=0\cdots(\mathrm{i})\\
\alpha^2-2\alpha-k=0\cdots(\mathrm{ii})
\end{array}
\]

(i)$-$(ii)より，$(k-4)\alpha+k-4=0$

したがって，$(k-4)(\alpha+1)=0$

ゆえに，$k=4$または$\alpha=-1$

（ア）$k=4$のとき

もとの2つの2次方程式は，ともに$x^2-2x-4=0$

よって，これを解くと，$x=1\pm\sqrt{5}$

これは，共通解がただ1つであることに反する．

（イ）$\alpha=-1$のとき

(i)に代入すると，$(-1)^2+(k-6)\cdot(-1)-4=0$

したがって，$k=3$

このとき，もとの2つの2次方程式は，$x^2-3x-4=0$と$x^2-2x-3=0$となり，解はそれぞれ，$x=-1,4$と$x=-1,3$

したがって，2つの方程式はただ1つの共通解$x=-1$をもつ．

よって，（ア），（イ）より，$k=3$，共通解は，$x=-1$

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% 問題I3.3.8：2次方程式の共通解（One More）★★★
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$x$についての2つの2次方程式$x^2+(k+3)x+8=0,x^2+5x+4k=0$が共通な実数解をもつとき，定数$k$の値と，そのときの共通解を求めよ．

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% 問題I3.3.8の解答（One More）★★★
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{共通解を$\alpha$として}，2つの2次方程式に$x=\alpha$を代入すると，

\[
\{\begin{array}{l}
\alpha^2+(k+3)\alpha+8=0\cdots(\mathrm{i})\\
\alpha^2+5\alpha+4k=0\cdots(\mathrm{ii})
\end{array}
\]

(i)$-$(ii)より，

\[
(k-2)\alpha+8-4k=0
\]

したがって，$(k-2)(\alpha-4)=0$

ゆえに，$k=2$または$\alpha=4$

（ア）$k=2$のとき

もとの2つの2次方程式は，ともに$x^2+5x+8=0$

この2次方程式の判別式を$D$とすると，$D=5^2-4\cdot 1\cdot 8=-7$

これは，$D<0$であるから，共通な実数解をもつことに反する．

（イ）$\alpha=4$のとき

(i)に代入すると，$4^2+(k+3)\cdot 4+8=0$

したがって，$k=-9$

このとき，もとの2つの2次方程式は，$x^2-6x+8=0$と$x^2+5x-36=0$となり，解はそれぞれ，$x=4,2$と$x=4,-9$

したがって，2つの方程式は共通解$x=4$をもつ．

よって，（ア），（イ）より，$k=-9$，共通解は，$x=4$

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% 例題I3.3.8の側注：2次方程式の共通解（One More）★★★
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% 側注1
連立1次方程式のときのように，辺々を引くことで$\alpha,k$についての連立方程式を解く．なお，(ii)より$k=\alpha^2-2\alpha$を導いて，(i)に代入してもよいが，3次方程式となり数学Iの範囲では解くことができない．

% 側注2
共通な解が2つとなる．

% 側注3
(i)か(ii)に代入する．

% 側注4
解が2つとも一致していない（ただ1つの共通解である）ことを確認する．

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% 問題I3.3.8の側注：2次方程式の共通解（One More）★★★
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% 側注1
実数解をもたない．

% 側注2
(i)か(ii)に代入する．

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% 例題I3.3.8の考え方・補足など：2次方程式の共通解（One More）★★★
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% 考え方
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ただ1つの共通解をもつので，それを$\alpha$とおくと扱いやすい．それぞれの方程式に代入すると，$\alpha,k$についての連立方程式と見ることができる．

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「$\alpha$とおく必要はないのではないか」という考え方もある．$x$のまま考えても構わないが，共通解の$x$であるか，共通解ではない$x$（方程式に含まれる未知数の$x$）であるかが混乱しやすい．そこで，共通解を$\alpha$とおくことを推奨する．

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解答では，2つの方程式$f(x)=0,g(x)=0$の共通解を$\alpha$とし，$\{\begin{array}{l}f(\alpha)=0\cdots(\mathrm{i})\\g(\alpha)=0\cdots(\mathrm{ii})\end{array}$が成り立ち，(i)$-$(ii)より，$f(\alpha)-g(\alpha)=0$を利用して，$\alpha$の値を求めている．

このことから類推して，「2つの方程式があり，それらの辺々を引き合わせてできた1次方程式の解が，2つの方程式の共通解なのではないか」と考える人もいるかもしれないが，これは成り立たないことがわかる．例えば，2つの方程式$x^2-2x-8=0,x^2-3x-18=0$について，共通解を$\alpha$とおくと，$\{\begin{array}{l}\alpha^2-2\alpha-8=0\cdots(\mathrm{i})\\\alpha^2-3\alpha-18=0\cdots(\mathrm{ii})\end{array}$が成り立ち，(i)$-$(ii)より，$\alpha+10=0$

よって，$\alpha=-10$が成り立つが，(i)，(ii)を解くと，解はそれぞれ$x=-2,4$と$x=-3,6$となり，実際には共通解をもたないことがわかる．