【数学I】例題3.3.16：2次不等式1（One More）★

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% 例題I3.3.16：2次不等式1（One More）★
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次の2次不等式を解け．

(1) $x^2-6x+5>0$

(2) $-2x^2+5x+3\geqq 0$

(3) $x^2-5x+2<0$

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% 例題I3.3.16の解答（One More）★
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(1) 左辺を因数分解すると，

\[
(x-1)(x-5)>0
\]

よって，不等式の解は，$x<1$または$5<x$

(2) 両辺に$-1$を掛けて，

\[
2x^2-5x-3\leqq 0
\]

左辺を因数分解すると，

\[
(2x+1)(x-3)\leqq 0
\]

よって，不等式の解は，$-\frac{1}{2}\leqq x\leqq 3$

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% 例題I3.3.16の別解
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左辺を因数分解すると，

\[
-(2x+1)(x-3)\geqq 0
\]

よって，$-\frac{1}{2}\leqq x\leqq 3$

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% 例題I3.3.16の解答（One More）★
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(3) $x^2-5x+2=0$を解くと，$x=\frac{5\pm\sqrt{17}}{2}$

よって，不等式の解は，

\[
\frac{5-\sqrt{17}}{2}<x<\frac{5+\sqrt{17}}{2}
\]

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% 問題I3.3.16：2次不等式1（One More）★
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次の2次不等式を解け．

(1) $x^2-8x+12>0$

(2) $x^2-3x+1>0$

(3) $-3x^2+5x+2\geqq 0$

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% 問題I3.3.16の解答（One More）★
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(1) 左辺を因数分解すると，

\[
(x-2)(x-6)>0
\]

よって，不等式の解は，$x<2$または$6<x$

(2) $x^2-3x+1=0$を解くと，$x=\frac{3\pm\sqrt{5}}{2}$

よって，不等式の解は，$x<\frac{3-\sqrt{5}}{2}$または$\frac{3+\sqrt{5}}{2}<x$

(3) 両辺に$-1$を掛けて，

\[
3x^2-5x-2\leqq 0
\]

左辺を因数分解すると，

\[
(3x+1)(x-2)\leqq 0
\]

よって，不等式の解は，$-\frac{1}{3}\leqq x\leqq 2$

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% 問題I3.3.16の別解
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左辺を因数分解すると，

\[
-(3x+1)(x-2)\geqq 0
\]

よって，$-\frac{1}{3}\leqq x\leqq 2$

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% 例題I3.3.16の側注：2次不等式1（One More）★
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% 側注1
$y=x^2-6x+5$は$x$軸と$x=1,5$で共有点をもつことがわかる．

% 側注2
両辺に$-1$を掛けると，不等式が扱いやすい形に変形できる．なお，不等号の向きが変わることに注意すること．

% 側注3
$-2x^2+5x+3\geqq 0$のグラフをかいて求めてもよい．

% 側注4
解の公式を用いる．
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]

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% 問題I3.3.16の側注：2次不等式1（One More）★
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% 側注1
$y=x^2-8x+12$は$x$軸と$x=2,6$で共有点をもつことがわかる．

% 側注2
解の公式を用いる．
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]

% 側注3
$-3x^2+5x+2\geqq 0$のグラフをかいて求めてもよい．

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% 例題I3.3.16の考え方・補足など：2次不等式1（One More）★
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% 考え方
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2次関数のグラフをかいて，グラフが$x$軸より上側にある$x$の範囲や，下側にある$x$の範囲を求める．すると，$x$軸との共有点から，不等式の解がわかる．なお，不等式を扱いやすくするために，$x^2$の係数は正にするとよい．

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% 余談
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$A,B$について，$AB>0\Longleftrightarrow\{\begin{array}{l}A>0\\B>0\end{array}$または$\{\begin{array}{l}A<0\\B<0\end{array}$が成り立つ（$A,B$は実数）．(1)において，$x-1$と$x-5$の符号をまとめると右の表のようになり，$x<1$または$5<x$で$(x-1)(x-5)>0$となることがわかる．

\begin{tabular}{|c|c|c|c|c|c|}
\hline$x$&$\cdots$&$1$&$\cdots$&$5$&$\cdots$\\

\hline$x-1$&$-$&$0$&$+$&$+$&$+$\\

\hline$x-5$&$-$&$-$&$-$&$0$&$+$\\

\hline$(x-1)(x-5)$&$+$&$0$&$-$&$0$&$+$\\

\hline
\end{tabular}