【数学I】例題3.3.6：2次方程式が実数解をもつ条件1（One More）★★

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% 例題I3.3.6：2次方程式が実数解をもつ条件1（One More）★★
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次の問いに答えよ．ただし，$k$を定数とする．

(1) $x$についての2次方程式$x^2-(k-3)x-3k+1=0$が重解をもつような$k$の値を定めよ．また，そのときの解を求めよ．

(2) $x$についての2次方程式$x^2+2kx+k^2-3k+6=0$の実数解の個数を調べよ．

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% 例題I3.3.6の解答（One More）★★
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(1) 与えられた2次方程式の判別式を$D$とすると，

\[
D=\{-(k-3)\}^2-4\cdot 1\cdot(-3k+1)=k^2+6k+5=(k+5)(k+1)
\]

2次方程式が重解をもつから，$D=0$

したがって，$k=-5,-1$

また，重解は，$x=-\frac{-(k-3)}{2\cdot 1}=\frac{k-3}{2}$

よって，

\begin{align*}
&k=-5\text{のとき，}&&\text{解は，}x=-4\\
&k=-1\text{のとき，}&&\text{解は，}x=-2
\end{align*}

(2) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(2k)^2-4\cdot 1\cdot(k^2-3k+6)=12k-24
\]

よって，実数解の個数は，$k$の値によって次のようになる．

$D>0$，すなわち，$k>2$のとき，$2$個

$D=0$，すなわち，$k=2$のとき，$1$個

$D<0$，すなわち，$k<2$のとき，$0$個

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% 問題I3.3.6：2次方程式が実数解をもつ条件1（One More）★★
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次の問いに答えよ．ただし，$k$を定数とする．

(1) $x$についての2次方程式$x^2+2kx+3k+10=0$が重解をもつような$k$の値を定めよ．また，そのときの解を求めよ．

(2) $x$についての2次方程式$x^2-2kx+k^2+4k-8=0$の実数解の個数を調べよ．

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% 問題I3.3.6の解答（One More）★★
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(1) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(2k)^2-4\cdot 1\cdot(3k+10)=4k^2-12k-40=4(k^2-3k-10)
\]

2次方程式が重解をもつから，$D=0$

したがって，$k=5,-2$

また，重解は，$x=-\frac{2k}{2\cdot 1}=-k$

よって，

\begin{align*}
&k=5\text{のとき，}&&\text{解は，}x=-5\\
&k=-2\text{のとき，}&&\text{解は，}x=2
\end{align*}

(2) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(-2k)^2-4\cdot 1\cdot(k^2+4k-8)=-16k+32
\]

よって，実数解の個数は，$k$の値によって次のようになる．

$D>0$，すなわち，$k<2$のとき，$2$個

$D=0$，すなわち，$k=2$のとき，$1$個

$D<0$，すなわち，$k>2$のとき，$0$個

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% 例題I3.3.6の側注：2次方程式が実数解をもつ条件1（One More）★★
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% 側注1
$D=b^2-4ac$を考える．$x$の係数は$-(k-3)$，定数項は$-3k+1$より，$b=-(k-3),c=-3k+1$

% 側注2
2次方程式$ax^2+bx+c=0$の重解は，$x=-\frac{b}{2a}$である．なお，重解は$x^2-(k-3)x-3k+1=0$に，$k=-5,-1$をそれぞれ代入することでも求めることができる．

% 側注3
$ax^2+bx+c=0$における$b=2b^\prime$（$x$の係数が2の倍数）のときであるので，$\frac{D}{4}=b^{\prime 2}-ac$を用いてもよい．

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% 問題I3.3.6の側注：2次方程式が実数解をもつ条件1（One More）★★
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% 側注1
$ax^2+bx+c=0$における$b=2b^\prime$（$x$の係数が2の倍数）のときであるので，$\frac{D}{4}=b^{\prime 2}-ac$を用いてもよい．

% 側注2
$ax^2+bx+c=0$における$b=2b^\prime$（$x$の係数が2の倍数）のときであるので，$\frac{D}{4}=b^{\prime 2}-ac$を用いてもよい．

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% 例題I3.3.6の考え方・補足など：2次方程式が実数解をもつ条件1（One More）★★
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% 考え方
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(1) 2次方程式$ax^2+bx+c=0$が重解をもつとき，その重解は，$x=-\frac{b}{2a}$であることを用いる．

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% 2次方程式の解の個数
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2次方程式$ax^2+bx+c=0$の判別式を$D=b^2-4ac$とすると，

\begin{align*}
&D>0\Longleftrightarrow\text{異なる2つの実数解をもつ．}\\
&D=0\Longleftrightarrow\text{1つの実数解(重解)をもつ．}\\
&D<0\Longleftrightarrow\text{実数解をもたない．}
\end{align*}