【数学I】例題3.3.30：2次方程式が実数解をもつ条件4（One More）★★★

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% 例題I3.3.30：2次方程式が実数解をもつ条件4（One More）★★★
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2つの2次方程式$x^2-2ax+a+2=0,x^2+ax+a^2+3a=0$について，次の条件を満たすような定数$a$の値の範囲を求めよ．

(1) 2つの方程式がともに実数解をもつ．

(2) 2つの方程式の少なくとも一方が実数解をもつ．

(3) 2つの方程式どちらか一方のみが実数解をもつ．

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% 例題I3.3.30の解答（One More）★★★
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2次方程式$x^2-2ax+a+2=0$の判別式を$D_1$とすると，

\[
\frac{D_1}{4}=(-a)^2-1\cdot(a+2)=a^2-a-2=(a+1)(a-2)
\]

2次方程式$x^2+ax+a^2+3a=0$の判別式を$D_2$とすると，

\[
D_2=a^2-4\cdot 1\cdot(a^2+3a)=-3a^2-12a=-3a(a+4)
\]

(1) 2つの方程式がともに実数解をもつのは，$D_1\geqq 0$かつ$D_2\geqq 0$のときである．

$D_1\geqq 0$であるから，$(a+1)(a-2)\geqq 0$

したがって，$a\leqq-1,2\leqq a\cdots(\mathrm{i})$

$D_2\geqq 0$であるから，$-3a(a+4)\geqq 0$

ゆえに，$-4\leqq a\leqq 0\cdots(\mathrm{ii})$

よって，(i)と(ii)の共通範囲を求めると，$-4\leqq a\leqq-1$

(2) 2つの方程式の少なくとも一方が実数解をもつのは，$D_1\geqq 0$または$D_2\geqq 0$のときである．

(i)，(ii)より，$a\leqq 0,2\leqq a$

(3) 2つの方程式のどちらか一方のみが実数解をもつのは，$D_1\geqq 0,D_2\geqq 0$の一方のみが成り立つことである．

(i)と(ii)の一方のみが成り立つ$a$の値の範囲を求めると，$a<-4,-1<a\leqq 0,2\leqq a$

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% 問題I3.3.30：2次方程式が実数解をもつ条件4（One More）★★★
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2つの2次方程式$x^2-6x+a^2=0,x^2-2(a-1)x-a^2-10a+1=0$について，次の条件を満たすような定数$a$の値の範囲を求めよ．

(1) 2つの方程式がともに実数解をもつ．

(2) 2つの方程式の少なくとも一方が実数解をもつ．

(3) 2つの方程式のどちらか一方のみが実数解をもつ．

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% 問題I3.3.30の解答（One More）★★★
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2次方程式$x^2-6x+a^2=0$の判別式を$D_1$とすると，

\[
\frac{D_1}{4}=(-3)^2-1\cdot a^2=9-a^2=-(a+3)(a-3)
\]

2次方程式$x^2-2(a-1)x-a^2-10a+1=0$の判別式を$D_2$とすると，

\[
\frac{D_2}{4}=\{-(a-1)\}^2-1\cdot(-a^2-10a+1)=2a^2+8a=2a(a+4)
\]

(1) 2つの方程式がともに実数解をもつのは，$D_1\geqq 0$かつ$D_2\geqq 0$のときである．

$D_1\geqq 0$であるから，$-(a+3)(a-3)\geqq 0$

したがって，$-3\leqq a\leqq 3\cdots(\mathrm{i})$

$D_2\geqq 0$であるから，$2a(a+4)\geqq 0$

ゆえに，$a\leqq-4,0\leqq a\cdots(\mathrm{ii})$

よって，(i)と(ii)の共通範囲を求めると，$0\leqq a\leqq 3$

(2) 2つの方程式の少なくとも一方が実数解をもつのは，$D_1\geqq 0$または$D_2\geqq 0$のときである．

(i)，(ii)より，$a\leqq-4,-3\leqq a$

(3) 2つの方程式のどちらか一方のみが実数解をもつのは，$D_1\geqq 0,D_2\geqq 0$の一方のみが成り立つことである．

(i)と(ii)の一方のみが成り立つ$a$の値の範囲を求めると，$a\leqq-4,-3\leqq a<0,3<a$

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% 例題I3.3.30の側注：2次方程式が実数解をもつ条件4（One More）★★★
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% 側注1
(i)，(ii)のどちらも実数解をもたないとき，すなわち，$D_1<0$かつ$D_2<0$を用いて，それ以外の範囲を考えることでも求めることができる．

% 側注2
2つの判別式を考えるので，$D_1,D_2$とする．

% 側注3
実数解をもつ条件は，$D>0,D=0$を合わせた，$D\geqq 0$である．

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% 問題I3.3.30の側注：2次方程式が実数解をもつ条件4（One More）★★★
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% 側注1
2つの判別式を考えるので，$D_1,D_2$とする．

% 側注2
実数解をもつ条件は，$D>0,D=0$を合わせた，$D\geqq 0$である．

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% 例題I3.3.30の考え方・補足など：2次方程式が実数解をもつ条件4（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，実数解をもつとき，$D\geqq 0$であることを利用する．

(1) 「ともに実数解をもつ」は，解の共通範囲である，$D_1\geqq 0$かつ$D_2\geqq 0$を求めればよい．

(2) 「少なくとも一方」は，解を合わせた範囲である，$D_1\geqq 0$または$D_2\geqq 0$を求めればよい．

(3) 「どちらか一方」は，$D_1\geqq 0,D_2\geqq 0$の一方のみが成り立つことであり，(2)から(1)を除いた範囲である．