【数学I】例題3.3.18：連立2次不等式（One More）★★

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% 例題I3.3.18：連立2次不等式（One More）★★
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次の連立不等式を解け．

(1) $\{\begin{array}{l}x^2+3x-10<0\\3x^2+4x+1>0\end{array}$

(2) $x^2-2\leqq 6x\leqq x^2-x$

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% 例題I3.3.18の解答（One More）★★
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(1) $x^2+3x-10<0$より，$(x+5)(x-2)<0$

したがって，$-5<x<2\cdots(\mathrm{i})$

$3x^2+4x+1>0$より，$(3x+1)(x+1)>0$

したがって，$x<-1,-\frac{1}{3}<x\cdots(\mathrm{ii})$

(i)，（ii）の共通範囲は，右の図のようになる．

よって，$-5<x<-1,-\frac{1}{3}<x<2$

(2) $x^2-2\leqq 6x$より，$x^2-6x-2\leqq 0$

$x^2-6x-2=0$を解くと，$x=3\pm\sqrt{11}$

したがって，$3-\sqrt{11}\leqq x\leqq 3+\sqrt{11}\cdots(\mathrm{i})$

$6x\leqq x^2-x$より，$x^2-7x\geqq 0$

因数分解すると，$x(x-7)\geqq 0$

したがって，$x\leqq 0,7\leqq x\cdots(\mathrm{ii})$

(i)，(ii)の共通範囲は，右の図のようになる．

よって，$3-\sqrt{11}\leqq x\leqq 0$

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% 問題I3.3.18：連立2次不等式（One More）★★
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次の連立不等式を解け．

(1) $\{\begin{array}{l}x^2+2x-3\leqq 0\\2x^2+3x+1>0\end{array}$

(2) $2x-1\leqq x^2-1\leqq 2x+3$

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% 問題I3.3.18の解答（One More）★★
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(1) $x^2+2x-3\leqq 0$より，$(x+3)(x-1)\leqq 0$

したがって，$-3\leqq x\leqq 1\cdots(\mathrm{i})$

$2x^2+3x+1>0$より，$(2x+1)(x+1)>0$

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

(i)，(ii)の共通範囲は，右の図のようになる．

よって，$-3\leqq x<-1,-\frac{1}{2}<x\leqq 1$

(2) $2x-1\leqq x^2-1$より，$x^2-2x\geqq 0$

因数分解すると，$(x-2)x\geqq 0$

したがって，$x\leqq 0,x\geqq 2\cdots(\mathrm{i})$

$x^2-1\leqq 2x+3$より，$x^2-2x-4\leqq 0$

$x^2-2x-4=0$を解くと，$x=1\pm\sqrt{5}$

したがって，$1-\sqrt{5}\leqq x\leqq 1+\sqrt{5}\cdots(\mathrm{ii})$

(i)，(ii)の共通範囲は，右の図のようになる．

よって，$1-\sqrt{5}\leqq x\leqq 0,2\leqq x\leqq 1+\sqrt{5}$

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% 例題I3.3.18の側注：連立2次不等式（One More）★★
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% 側注1
たすき掛けを用いる．

% 側注2
$3^2<(\sqrt{11})^2<4^2$より，$3<\sqrt{11}<4$であることから，$3-\sqrt{11}<0,3+\sqrt{11}<7$がわかる．

% 側注3
$A\leqq B\leqq C$は連立不等式

\[
\{\begin{array}{l}
A\leqq B\\
B\leqq C
\end{array}
\]

と同じ意味である．

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% 問題I3.3.18の側注：連立2次不等式（One More）★★
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% 側注1
たすき掛けを用いる．

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% 例題I3.3.18の考え方・補足など：連立2次不等式（One More）★★
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% 考え方
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それぞれの不等式を解き，それらの共通範囲を求める．共通範囲を求めるには，連立1次不等式のときと同じように，数直線を用いて解を図示するとよい．

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数直線上の共通範囲が連立不等式の解となる．