【数学I】例題3.3.11：x軸から切り取る線分の長さ（One More）★★

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% 例題I3.3.11：$x$軸から切り取る線分の長さ（One More）★★
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(1) 2次関数$y=-3x^2-5x+5$のグラフが$x$軸から切り取る線分の長さを求めよ．

(2) 放物線$y=x^2+2x+2k$が$x$軸から切り取る線分の長さが4であるとき，定数$k$の値を求めよ．

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% 例題I3.3.11の解答（One More）★★
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(1) $-3x^2-5x+5=0$とすると，$3x^2+5x-5=0$

したがって，$x=\frac{-5\pm\sqrt{5^2-4\cdot 3\cdot(-5)}}{2\cdot 3}=\frac{-5\pm\sqrt{25+60}}{6}=\frac{-5\pm\sqrt{85}}{6}$

よって，放物線が$x$軸から切り取る線分の長さは，

\[
\frac{-5+\sqrt{85}}{6}-\frac{-5-\sqrt{85}}{6}=\frac{\sqrt{85}}{3}
\]

(2) $x^2+2x+2k=0\cdots(\mathrm{i})$とする．この2次方程式の判別式を$D$とすると，

\[
\frac{D}{4}=1^2-1\cdot 2k=1-2k
\]

グラフが$x$軸と異なる2点で交わるから，$D>0$

したがって，$1-2k>0$より，$k<\frac{1}{2}\cdots(\mathrm{ii})$

このとき，2次方程式(i)を解くと，$x=-1\pm\sqrt{1-2k}$

切り取る線分の長さが4であるから，

\[
(-1+\sqrt{1-2k})-(-1-\sqrt{1-2k})=4
\]

ゆえに，$\sqrt{1-2k}=2$

したがって，$1-2k=4$

これより，$k=-\frac{3}{2}$

これは，(ii)を満たす．

よって，$k=-\frac{3}{2}$

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% 問題I3.3.11：$x$軸から切り取る線分の長さ（One More）★★
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(1) 2次関数$y=-2x^2+3x+4$のグラフが$x$軸から切り取る線分の長さを求めよ．

(2) 放物線$y=-x^2+4x+2k$が$x$軸から切り取る線分の長さが6であるとき，定数$k$の値を求めよ．

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% 問題I3.3.11の解答（One More）★★
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(1) $-2x^2+3x+4=0$とすると，$2x^2-3x-4=0$

したがって，$x=\frac{3\pm\sqrt{(-3)^2-4\cdot 2\cdot(-4)}}{2\cdot 2}=\frac{3\pm\sqrt{9+32}}{4}=\frac{3\pm\sqrt{41}}{4}$

よって，放物線が$x$軸から切り取る線分の長さは，

\[
\frac{3+\sqrt{41}}{4}-\frac{3-\sqrt{41}}{4}=\frac{\sqrt{41}}{2}
\]

(2) $-x^2+4x+2k=0$とすると，$x^2-4x-2k=0\cdots(\mathrm{i})$

この2次方程式の判別式を$D$とすると，

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

グラフが$x$軸と異なる2点で交わるから，$D>0$

したがって，$4+2k>0$より，$k>-2\cdots(\mathrm{ii})$

このとき，2次方程式(i)を解くと，

\[
x=2\pm\sqrt{4+2k}
\]

切り取る線分の長さが6であるから，

\[
(2+\sqrt{4+2k})-(2-\sqrt{4+2k})=6
\]

ゆえに，$2\sqrt{4+2k}=6$

したがって，$4+2k=9$

これより，$k=\frac{5}{2}$

この値は，(ii)を満たす．

よって，求める定数$k$の値は，$\frac{5}{2}$

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% 例題I3.3.11の側注：$x$軸から切り取る線分の長さ（One More）★★
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% 側注1
扱いやすさのために，$x^2$の係数を正にする．

% 側注2
$x$軸と異なる2点で交わらなければ，線分ができない．

% 側注3
$\sqrt{1-2k}>0$より，$-1+\sqrt{1-2k}$の方が$-1-\sqrt{1-2k}$より大きい．

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% 問題I3.3.11の側注：$x$軸から切り取る線分の長さ（One More）★★
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% 側注1
扱いやすさのために，$x^2$の係数を正にする．

% 側注2
扱いやすさのために，$x^2$の係数を正にする．

% 側注3
$x$軸と異なる2点で交わらなければ，線分ができない．

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% 例題I3.3.11の考え方・補足など：$x$軸から切り取る線分の長さ（One More）★★
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% 考え方
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「$x$軸から切り取る線分の長さ」とは，放物線$y=ax^2+bx+c$が$x$軸と異なる2点$\mathrm{A},\mathrm{B}$で交わるときの線分ABの長さのことである．2点A，Bの$x$座標をそれぞれ$\alpha,\beta(\alpha<\beta)$とすると，$\mathrm{AB}=\beta-\alpha$である．

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% 余談
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放物線$y=ax^2+bx+c$が$x$軸と異なる2点$\mathrm{A},\mathrm{B}$で交わるとき，2次方程式$ax^2+bx+c=0$の判別式を$D$とすると，$D>0$であり，線分ABの長さは，
\[
\mathrm{AB}=\frac{\sqrt{b^2-4ac}}{|a|}=\frac{\sqrt{D}}{|a|}
\]
で表されることが知られている．