【数学I】例題3.3.3：方程式の解（係数が文字のとき）（One More）★★★

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% 例題I3.3.3：方程式の解（係数が文字のとき）（One More）★★★
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次の方程式を解け．ただし，$a$は定数とする．

(1) $ax^2+(a+2)x+2=0$

(2) $(a^2-a)x^2=a-1$

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% 例題I3.3.3の解答（One More）★★★
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(1)

(i) $a=0$のとき

方程式は，$2x+2=0$より，$x=-1$

(ii) $a\neq 0$のとき

$ax^2+(a+2)x+2=0$より，$(x+1)(ax+2)=0$

したがって，$x=-1,-\frac{2}{a}$

よって，(i)，(ii)より，求める解は，

\[
\begin{cases}a=0\text{のとき，}&x=-1\\a\neq 0\text{のとき，}&x=-1,-\frac{2}{a}\end{cases}
\]

(2)

(i) $a=0$のとき

方程式は，$0\cdot x^2=0-1$

これを満たす$x$は存在しないので，解なし

(ii) $a=1$のとき

方程式は，$0\cdot x^2=0$

これは，$x$の値に関わらず成り立つ．

したがって，解はすべての実数

(iii) $a\neq 0,1$のとき

$a^2-a\neq 0$から，両辺を$a^2-a$で割ると，$x^2=\frac{1}{a}$

（ア）$a>0$のとき，$x=\pm\sqrt{\frac{1}{a}}=\pm\frac{\sqrt{a}}{a}$

（イ）$a<0$のとき，解なし

よって，(i)〜(iii)より，求める解は，

\[
\begin{cases}a\leqq 0\text{のとき，}&\text{解なし}\\a=1\text{のとき，}&\text{解はすべての実数}\\0<a<1,1<a\text{のとき，}&x=\pm\frac{\sqrt{a}}{a}
\end{cases}
\]

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% 問題I3.3.3：方程式の解（係数が文字のとき）（One More）★★★
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次の方程式を解け．ただし，$a$は定数とする．

(1) $ax^2+(a+3)x+3=0$

(2) $(a^2-2a)x^2=a-2$

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% 問題I3.3.3の解答（One More）★★★
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(1)

(i) $a=0$のとき

方程式は，$3x+3=0$より，$x=-1$

(ii) $a\neq 0$のとき

$ax^2+(a+3)x+3=0$より，$(x+1)(ax+3)=0$

したがって，$x=-1,-\frac{3}{a}$

よって，(i)，(ii)より，求める解は，

\[
\begin{cases}a=0\text{のとき，}&x=-1\\a\neq 0\text{のとき，}&x=-1,-\frac{3}{a}\end{cases}
\]

(2)

(i) $a=0$のとき

方程式は，$0\cdot x^2=0-2$

これを満たす$x$は存在しないので，解なし

(ii) $a=2$のとき

方程式は，$0\cdot x^2=0$

これは，$x$の値に関わらず成り立つ．

したがって，解はすべての実数

(iii) $a\neq 0,2$のとき

$a^2-2a\neq 0$から，両辺を$a^2-2a$で割ると，$x^2=\frac{1}{a}$

（ア）$a>0$のとき，$x=\pm\sqrt{\frac{1}{a}}=\pm\frac{1}{\sqrt{a}}=\pm\frac{\sqrt{a}}{a}$

（イ）$a<0$のとき，解なし

よって，(i)〜(iii)より，求める解は，

\[
\begin{cases}a\leqq 0\text{のとき，}&\text{解なし}\\a=2\text{のとき，}&\text{解はすべての実数}\\0<a<2,2<a\text{のとき，}&x=\pm\frac{1}{\sqrt{a}}
\end{cases}
\]

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% 例題I3.3.3の側注：方程式の解（係数が文字のとき）（One More）★★★
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% 側注1
$a=0$のとき，$x^2$の項がなくなるので，$x$の1次方程式になる．

% 側注2
たすき掛けを用いる．

% 側注3
$a-1=0$の可能性があり，両辺を単に$a-1$で割ってしまうと，$0$で割ってしまうことになる．そこで，場合分けを考える．

% 側注4
$x^2=\frac{a-1}{a^2-a}=\frac{a-1}{a(a-1)}$

% 側注5
$x^2=\frac{1}{a}$を解く．

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% 問題I3.3.3の側注：方程式の解（係数が文字のとき）（One More）★★★
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% 側注1
$a=0$のとき，$x^2$の項がなくなるので，$x$の1次方程式になる．

% 側注2
たすき掛けを用いる．

% 側注3
$x^2=\frac{a-2}{a(a-2)}=\frac{1}{a}$

% 側注4
$x^2=\frac{1}{a}$を解く．

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% 例題I3.3.3の考え方・補足など：方程式の解（係数が文字のとき）（One More）★★★
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問題文では単に「方程式」となっていることから，2次方程式であるとは限らない．最高次の項の係数が$0$のときを考えて，場合分けをして解けばよい．

(1) $a=0$のとき，$0\cdot x^2+(0+2)x+2=0$となり，1次方程式となる．$a\neq 0$のとき，そのまま，$ax^2+(a+2)x+2=0$となり，2次方程式となる．

(2) (1)と同様に，最高次の項$x^2$の係数が$0$のときを考えて，場合分けをする．なお，$a-1=0$の可能性があるので，両辺を単に$a-1$で割ってしまわないように注意すること（場合分けを考える）．