【数学I】例題3.3.17：2次不等式2（One More）★

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% 例題I3.3.17：2次不等式2（One More）★
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次の2次不等式を解け．

(1) $x^2-8x+16\leqq 0$

(2) $-3x^2+6x-3<0$

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

(4) $-x^2+2x-4>0$

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% 例題I3.3.17の解答（One More）★
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(1) 左辺を因数分解すると，

\[
(x-4)^2\leqq 0
\]

よって，不等式の解は，$x=4$

(2) $-3x^2+6x-3<0$より，$3x^2-6x+3>0$

因数分解すると，$3(x-1)^2>0$

よって，不等式の解は，$1\text{以外のすべての実数}$

(3) 2次方程式$x^2-2x+3=0$の判別式を$D$とすると，

\[
\frac{D}{4}=(-1)^2-1\cdot 3=-2
\]

よって，$D<0$であるから，不等式の解は，すべての実数

(4) $-x^2+2x-4>0$より，$x^2-2x+4<0$

2次方程式$x^2-2x+4=0$の判別式を$D$とすると，

\[
\frac{D}{4}=(-1)^2-1\cdot 4=-3
\]

よって，$D<0$であるから，不等式の解はない

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% 問題I3.3.17：2次不等式2（One More）★
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次の2次不等式を解け．

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

(2) $x^2+4x+5>0$

(3) $x^2-6x+9\leqq 0$

(4) $-x^2+4x-4>0$

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% 問題I3.3.17の解答（One More）★
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(1) $-2x^2+4x-2<0$より，$2x^2-4x+2>0$

因数分解すると，$2(x-1)^2>0$

よって，不等式の解は，$1\text{以外のすべての実数}$

(2) 2次方程式$x^2+4x+5=0$の判別式を$D$とすると，

\[
D=4^2-4\cdot 1\cdot 5=16-20=-4
\]

よって，$D<0$であるから，不等式の解は，すべての実数

(3) 左辺を因数分解すると，

\[
(x-3)^2\leqq 0
\]

よって，不等式の解は，$x=3$

(4) $-x^2+4x-4>0$より，$x^2-4x+4<0$

左辺を因数分解すると，$(x-2)^2<0$

よって，不等式の解はない

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% 例題I3.3.17の側注：2次不等式2（One More）★
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% 側注1
2次関数$y=x^2-8x+16$は，$x=4$のとき，$y=0$であり，それ以外のときは$y>0$となるから，$y\leqq 0$となる$x$の値は$x=4$のみである．

% 側注2
2次関数$y=3x^2-6x+3$は，$x=1$のとき，$y=0$となるから，$x=1$を除く．なお，$y=-3x^2+6x-3$を考えてもよい．

% 側注3
$D<0$より，グラフは$x$軸と共有点をもたない．なお，グラフは常に$y>0$が成り立っており，不等号の向きを混同しないように注意すること．

% 側注4
グラフが$x$軸より下側となる部分がない．

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% 問題I3.3.17の側注：2次不等式2（One More）★
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% 側注1
2次関数$y=2x^2-4x+2$は，$x=1$のとき，$y=0$となるから，$x=1$を除く．なお，$y=-2x^2+4x-2$を考えてもよい．

% 側注2
2次関数$y=x^2-6x+9$は，$x=3$のとき，$y=0$であり，それ以外のときは$y>0$となるから，$y\leqq 0$となる$x$の値は$x=3$のみである．

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% 例題I3.3.17の考え方・補足など：2次不等式2（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，不等式の解を求める．グラフと$x$軸との共有点があるか否かを判断するには，2次不等式の不等号を等号におき換えた，2次方程式$ax^2+bx+c=0$の判別式$D$の符号を考えればよい．

$D=0$のとき

$D<0$のとき