【数学I】例題3.3.5：実数解の個数と判別式（One More）★

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% 例題I3.3.5：実数解の個数と判別式（One More）★
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次の2次方程式の実数解の個数を調べよ．

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

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

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

(4) $9x^2-30x+25=0$

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% 例題I3.3.5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，

\[
D=(-1)^2-4\cdot 5\cdot 2=-39
\]

$D<0$であるから，実数解の個数は，$0$個

(2) 与えられた2次方程式の判別式を$D$とすると，

\[
D=2^2-4\cdot 1\cdot(-2)=12
\]

$D>0$であるから，実数解の個数は，$2$個

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% 例題I3.3.5の別解
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与えられた2次方程式の判別式を$D$とすると，$\frac{D}{4}=1^2-1\cdot(-2)=3$

$D>0$であるから，実数解の個数は，$2$個

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% 例題I3.3.5の解答（One More）★
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(3) 与えられた2次方程式の判別式を$D$とすると，

\[
D=0^2-4\cdot 3\cdot(-4)=48
\]

$D>0$であるから，実数解の個数は，$2$個

(4) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(-30)^2-4\cdot 9\cdot 25=900-900=0
\]

$D=0$であるから，実数解の個数は，$1$個

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% 例題I3.3.5の別解
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与えられた2次方程式の判別式を$D$とすると，$\frac{D}{4}=(-15)^2-9\cdot 25=0$

$D=0$であるから，実数解の個数は，$1$個

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% 問題I3.3.5：実数解の個数と判別式（One More）★
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次の2次方程式の実数解の個数を調べよ．

(1) $x^2-4x+1=0$

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

(3) $2x^2-6=0$

(4) $16x^2-8x+1=0$

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% 問題I3.3.5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，

\[
D=(-4)^2-4\cdot 1\cdot 1=12
\]

$D>0$であるから，実数解の個数は，$2$個

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% 問題I3.3.5の別解
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与えられた2次方程式の判別式を$D$とすると，$\frac{D}{4}=(-2)^2-1\cdot 1=3$

$D>0$であるから，実数解の個数は，$2$個

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% 問題I3.3.5の解答（One More）★
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(2) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(2)^2-4\cdot 4\cdot 3=-44
\]

$D<0$であるから，実数解の個数は，$0$個

(3) 与えられた2次方程式の判別式を$D$とすると，

\[
D=0^2-4\cdot 2\cdot(-6)=48
\]

$D>0$であるから，実数解の個数は，$2$個

(4) 与えられた2次方程式の判別式を$D$とすると，

\[
D=(-8)^2-4\cdot 16\cdot 1=64-64=0
\]

$D=0$であるから，実数解の個数は，$1$個

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% 問題I3.3.5の別解
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与えられた2次方程式の判別式を$D$とすると，$\frac{D}{4}=(-4)^2-16\cdot 1=0$

$D=0$であるから，実数解の個数は，$1$個

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% 例題I3.3.5の側注：実数解の個数と判別式（One More）★
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% 側注1
$ax^2+bx+c=0$において，判別式を$D$とすると，
\[
D=b^2-4ac
\]

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

% 側注3
$\frac{D}{4}=b^{\prime 2}-ac$

% 側注4
$3x^2+0\cdot x-4=0$

% 側注5
\begin{align*}
&9x^2-30x+25\\
=&(3x-5)^2
\end{align*}

% 側注6
重解である．

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% 問題I3.3.5の側注：実数解の個数と判別式（One More）★
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% 側注1
$ax^2+bx+c=0$において$b=2b^\prime$（$x$の係数が2の倍数）のとき，$\frac{D}{4}=b^{\prime 2}-ac$を用いてもよい．

% 側注2
$2x^2+0\cdot x-6=0$

% 側注3
\begin{align*}
&16x^2-8x+1\\
=&(4x-1)^2
\end{align*}

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% 例題I3.3.5の考え方・補足など：実数解の個数と判別式（One More）★
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% 考え方
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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*}
&b^2-4ac>0(D>0)\Longleftrightarrow\text{異なる2つの実数解をもつ．}\\
&b^2-4ac=0(D=0)\Longleftrightarrow\text{1つの実数解(重解)をもつ．}\\
&b^2-4ac<0(D<0)\Longleftrightarrow\text{実数解をもたない．}
\end{align*}

とくに，$b=2b^{\prime}$であるとき，すなわち，$ax^2+2b^\prime x+c=0$のとき，$\frac{D}{4}=b^{\prime 2}-ac$を用いると計算が楽になる（別解）．

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% One Point
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2次方程式の実数解の個数は，判別式$D$の符号を考える．