【数学I】例題3.3.7：2次方程式が実数解をもつ条件2（One More）★★

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% 例題I3.3.7：2次方程式が実数解をもつ条件2（One More）★★
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$x$についての2つの2次方程式$3x^2+4x+k-3=0,x^2-(2k+3)x+k^2-2=0$がともに実数解をもつような定数$k$の値の範囲を求めよ．

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% 例題I3.3.7の解答（One More）★★
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2次方程式$3x^2+4x+(k-3)=0$の判別式を$D_1$とすると，

\[
D_1=4^2-4\cdot 3\cdot(k-3)=16-12k+36=52-12k
\]

2次方程式が実数解をもつから，$D_1\geqq 0$

したがって，$52-12k\geqq 0$

ゆえに，

\[
k\leqq\frac{13}{3}\cdots(\mathrm{i})
\]

2次方程式$x^2-(2k+3)x+(k^2-2)=0$の判別式を$D_2$とすると，

\[
D_2=\{-(2k+3)\}^2-4\cdot 1\cdot(k^2-2)=4k^2+12k+9-4k^2+8=12k+17
\]

2次方程式が実数解をもつから，$D_2\geqq 0$

したがって，$12k+17\geqq 0$

ゆえに，

\[
k\geqq-\frac{17}{12}\cdots(\mathrm{ii})
\]

よって，(i)と(ii)の共通範囲を求めると，

\[
-\frac{17}{12}\leqq k\leqq\frac{13}{3}
\]

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% 問題I3.3.7：2次方程式が実数解をもつ条件2（One More）★★
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$x$についての2つの2次方程式$x^2-2x-k+2=0,x^2+(2k-1)x+k^2+2=0$がともに実数解をもたないような定数$k$の値の範囲を求めよ．

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% 問題I3.3.7の解答（One More）★★
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2次方程式$x^2-2x-(k-2)=0$の判別式を$D_1$とすると，

\[
D_1=(-2)^2-4\cdot 1\cdot(-k+2)=4+4k-8=4k-4
\]

2次方程式が実数解をもたないから，$D_1<0$

したがって，$4k-4<0$

ゆえに，$k<1\cdots(\mathrm{i})$

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

\[
D_2=(2k-1)^2-4\cdot 1\cdot(k^2+2)=4k^2-4k+1-4k^2-8=-4k-7
\]

2次方程式が実数解をもたないから，$D_2<0$

したがって，$-4k-7<0$

ゆえに，$k>-\frac{7}{4}\cdots(\mathrm{ii})$

よって，(i)と(ii)の共通範囲を求めると，

\[
-\frac{7}{4}<k<1
\]

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% 例題I3.3.7の側注：2次方程式が実数解をもつ条件2（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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% 問題I3.3.7の側注：2次方程式が実数解をもつ条件2（One More）★★
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% 側注1
実数解をもたない条件は，$D<0$である．

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% 例題I3.3.7の考え方・補足など：2次方程式が実数解をもつ条件2（One More）★★
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% 考え方
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「2次方程式が実数解をもつ」とは，2つの条件を合わせた，
\[
\{\begin{array}{lr}\text{異なる2つの実数解をもつ．}&\Longleftrightarrow D>0\\\text{1つの実数解（重解）をもつ．}&\Longleftrightarrow D=0\end{array}
\]
である．したがって，2次方程式が実数解をもつ条件は，$D\geqq 0$と表される．

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% One Point
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2次方程式が実数解をもつ条件は，$D\geqq 0$