【数学I】例題3.3.1：2次方程式の解1（One More）★

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% 例題I3.3.1：2次方程式の解1（One More）★
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次の2次方程式を解け．

(1) $5x^2-11x-4=0$

(2) $3x^2-8x-5=0$

(3) $9x^2+6x+1=0$

(4) $8x^2-14x+3=0$

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% 例題I3.3.1の解答（One More）★
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(1) 解の公式より，

\[
x=\frac{-(-11)\pm\sqrt{(-11)^2-4\cdot 5\cdot(-4)}}{2\cdot 5}=\frac{11\pm\sqrt{121+80}}{10}=\frac{11\pm\sqrt{201}}{10}
\]

(2) 解の公式より，

\[
x=\frac{-(-4)\pm\sqrt{(-4)^2-3\cdot(-5)}}{3}=\frac{4\pm\sqrt{16+15}}{3}=\frac{4\pm\sqrt{31}}{3}
\]

(3) 左辺を因数分解すると，$(3x+1)^2=0$

よって，$3x+1=0$より，$x=-\frac{1}{3}$

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% 例題I3.3.1の別解
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\[
x=\frac{-3\pm\sqrt{3^2-9\cdot 1}}{9}=\frac{-3\pm\sqrt{9-9}}{9}=\frac{-3\pm 0}{9}=-\frac{1}{3}
\]

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% 例題I3.3.1の解答（One More）★
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(4) 左辺を因数分解すると，$(4x-1)(2x-3)=0$

したがって，$4x-1=0$または$2x-3=0$

よって，$x=\frac{1}{4},\frac{3}{2}$

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% 例題I3.3.1の別解
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\[
x=\frac{-(-7)\pm\sqrt{(-7)^2-8\cdot 3}}{8}=\frac{7\pm\sqrt{49-24}}{8}=\frac{7\pm\sqrt{25}}{8}=\frac{7\pm 5}{8}
\]

よって，$x=\frac{1}{4},\frac{3}{2}$

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% 問題I3.3.1：2次方程式の解1（One More）★
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次の2次方程式を解け．

(1) $4x^2-7x+2=0$

(2) $x^2-8x-5=0$

(3) $16x^2+8x+1=0$

(4) $6x^2-11x+3=0$

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% 問題I3.3.1の解答（One More）★
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(1) 解の公式より，

\[
x=\frac{-(-7)\pm\sqrt{(-7)^2-4\cdot 4\cdot 2}}{2\cdot 4}=\frac{7\pm\sqrt{49-32}}{8}=\frac{7\pm\sqrt{17}}{8}
\]

(2) 解の公式より，

\[
x=\frac{-(-4)\pm\sqrt{(-4)^2-1\cdot(-5)}}{1}=4\pm\sqrt{16+5}=4\pm\sqrt{21}
\]

(3) 左辺を因数分解すると，$(4x+1)^2=0$

よって，$4x+1=0$より，$x=-\frac{1}{4}$

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% 問題I3.3.1の別解
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\[
x=\frac{-8\pm\sqrt{8^2-4\cdot 16}}{32}=\frac{-8\pm\sqrt{64-64}}{32}=\frac{-8\pm 0}{32}=-\frac{1}{4}
\]

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% 問題I3.3.1の解答（One More）★
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(4) 左辺を因数分解すると，$(3x-1)(2x-3)=0$

したがって，$3x-1=0$または$2x-3=0$

よって，$x=\frac{1}{3},\frac{3}{2}$

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% 問題I3.3.1の別解
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\[
x=\frac{-(-11)\pm\sqrt{(-11)^2-4\cdot 6\cdot 3}}{2\cdot 6}=\frac{11\pm\sqrt{121-72}}{12}=\frac{11\pm\sqrt{49}}{12}=\frac{11\pm 7}{12}
\]

よって，$x=\frac{1}{3},\frac{3}{2}$

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% 例題I3.3.1の側注：2次方程式の解1（One More）★
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% 側注1
$a=5,b=-11,c=-4$として，
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
に代入する．

% 側注2
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注3
解の公式を用いてもよい．

% 側注4
2つの解が重なったものと考えて，このような解を重解という．

% 側注5
たすき掛けを用いる．

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% 問題I3.3.1の側注：2次方程式の解1（One More）★
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% 側注1
$a=4,b=-7,c=2$として，
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
に代入する．

% 側注2
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]

% 側注3
解の公式を用いてもよい．

% 側注4
たすき掛けを用いる．

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% 例題I3.3.1の考え方・補足など：2次方程式の解1（One More）★
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% 考え方
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2次方程式を解くには，因数分解するか，解の公式を用いるとよい．なお，因数分解はすべての方程式で上手くできるとは限らないが，解の公式より楽に解けることがある．そこで，先に因数分解を考え，上手くできなければ解の公式を用いることを推奨する．

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% 解の公式
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$ax^2+bx+c=0$の解は，$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

とくに，$ax^2+2b^{\prime}x+c=0$の解は，$x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}$