【数学A】例題2.1.5：2次方程式が満たす条件と確率（One More）★★★

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% 例題A2.1.5：2次方程式が満たす条件と確率（One More）★★★
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大中小3個のさいころを同時に投げ，出た目の数をそれぞれ$a,b,c$とするとき，$x$についての2次方程式$ax^2+bx+c=0$が重解をもつ確率を求めよ．

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% 例題A2.1.5の解答（One More）★★★
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2次方程式$ax^2+bx+c=0$の判別式を$D$とすると，$D=b^2-4ac$

2次方程式が重解をもつから，$D=0$より，$b^2-4ac=0$

$a,b,c$は整数であり，$b^2=4ac$より，$b$は偶数である．

(i) $b=2$のとき

$2^2=4ac$より，$ac=1$

したがって，$(a,c)=(1,1)$の1通り

(ii) $b=4$のとき

$4^2=4ac$より，$ac=4$

したがって，$(a,c)=(1,4),(2,2),(4,1)$の3通り

(iii) $b=6$のとき

$6^2=4ac$より，$ac=9$

したがって，$(a,c)=(3,3)$の1通り

よって，(i)〜(iii)より，求める確率は，

\[
\frac{1+3+1}{6^3}=\frac{5}{216}
\]

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% 問題A2.1.5：2次方程式が満たす条件と確率（One More）★★★
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大小2個のさいころを同時に投げ，出た目の数をそれぞれ$a,b$とするとき，$x$についての2次方程式$x^2+ax+b=0$が実数解をもつ確率を求めよ．

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% 問題A2.1.5の解答（One More）★★★
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2次方程式$x^2+ax+b=0$の判別式を$D$とすると，$D=a^2-4b$

2次方程式が実数解をもつから，$D\geqq 0$より，$a^2-4b\geqq 0$

したがって，$b\leqq\frac{1}{4}a^2$であるから，$b$と$\frac{1}{4}a^2$の大小を比較する．

$a,b$はそれぞれ1から6までの整数の値をとるから，

(i) $a=1$のとき

$b\leqq\frac{1}{4}$より，条件を満たす$b$はない．

(ii) $a=2$のとき

$b\leqq 1$より，$b=1$の1通り

(iii) $a=3$のとき

$b\leqq\frac{9}{4}$より，$b=1,2$の2通り

(iv) $a=4$のとき

$b\leqq 4$より，$b=1,2,3,4$の4通り

(v) $a=5$のとき$b\leqq\frac{25}{4}$より，$b=1,2,3,4,5,6$の6通り

(vi) $a=6$のとき

$b\leqq 9$より，$b=1,2,3,4,5,6$の6通り

よって，(i)〜(vi)より，求める確率は，$\frac{1+2+4+6+6}{6^2}=\frac{19}{36}$

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% 例題A2.1.5の側注：2次方程式が満たす条件と確率（One More）★★★
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% 側注1
さいころの目であるから，$a\neq 0$

% 側注2
$4ac$は偶数であるから，$b^2$は偶数である．これより，$b$は偶数である．

% 側注3
3個のさいころの目の出方は，$6^3$通り

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% 問題A2.1.5の側注：2次方程式が満たす条件と確率（One More）★★★
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% 側注1
$a=2,b=1$のとき，$x^2+ax+b=0$は重解をもつ．

% 側注2
$a=4,b=4$のとき，$x^2+ax+b=0$は重解をもつ．

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% 例題A2.1.5の考え方・補足など：2次方程式が満たす条件と確率（One More）★★★
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% 考え方
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2次方程式の判別式を$D$とすると，

\[
\text{2次方程式が重解をもつ．}\Longleftrightarrow b^2-4ac=0(D=0)
\]

$D=b^2-4ac$であることから，$b^2-4ac=0$となる$a,b,c$を求める．このとき，$a,b,c$はさいころの目であり，整数であることを利用して$a,b,c$の組を定める．