【数学I】例題3.3.4：2次方程式の係数決定（One More）★★

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% 例題I3.3.4：2次方程式の係数決定（One More）★★
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(1) 2次方程式$x^2+ax+b=0$の2つの解が$3$と$-2$であるとき，定数$a$，$b$の値を求めよ．

(2) 2次方程式$x^2+ax-32=0$の解の1つが$x=a$のとき，定数$a$の値を求めよ．また，そのときの他の解を求めよ．

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(1) $x^2+ax+b=0$の2つの解が$3$と$-2$であるから，$x=3$と$x=-2$をそれぞれ代入すると，

\[
\{\begin{array}{l}
3^2+a\cdot 3+b=0\\
(-2)^2+a\cdot(-2)+b=0
\end{array}
\]

すなわち，$\{\begin{array}{l}3a+b=-9\cdots(\mathrm{i})\\-2a+b=-4\cdots(\mathrm{ii)}\end{array}$

よって，(i)，(ii)を解いて，$a=-1,b=-6$

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2つの解が$3$と$-2$であるから，もとの2次方程式は，$(x-3)(x+2)=0$

したがって，$x^2-x-6=0$

よって，$x^2+ax+b=0$と係数を比較すると，$a=-1,b=-6$

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(2)

$x=a$が$x^2+ax-32=0$の解であるから，$a^2+a\cdot a-32=0$

すなわち，$2a^2=32$より，$a=\pm 4$

(i) $a=4$のとき

方程式は，$x^2+4x-32=0$

したがって，$(x-4)(x+8)=0$

ゆえに，$x=4,-8$

したがって，他の解は，$x=-8$

(ii) $a=-4$のとき

方程式は，$x^2-4x-32=0$

したがって，$(x-8)(x+4)=0$

ゆえに，$x=8,-4$

したがって，他の解は，$x=8$

よって，(i)，（ii）より，

\begin{align*}
&a=4\text{のとき，}&&\text{他の解は，}x=-8\\
&a=-4\text{のとき，}&&\text{他の解は，}x=8
\end{align*}

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% 問題I3.3.4：2次方程式の係数決定（One More）★★
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(1) 2次方程式$x^2+ax+b=0$の2つの解が$4$と$-5$であるとき，定数$a$，$b$の値を求めよ．

(2) 2次方程式$x^2+ax-8=0$の解の1つが$x=a$のとき，定数$a$の値を求めよ．また，そのときの他の解を求めよ．

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% 問題I3.3.4の解答（One More）★★
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(1) $x^2+ax+b=0$の2つの解が$4$と$-5$であるから，$x=4$と$x=-5$をそれぞれ代入して，

\[
\{\begin{array}{l}
4^2+a\cdot 4+b=0\\
(-5)^2+a\cdot(-5)+b=0
\end{array}
\]

すなわち，$\{\begin{array}{l}4a+b=-16\cdots(\mathrm{i})\\-5a+b=-25\cdots(\mathrm{ii})\end{array}$

よって，(i)，(ii)を解いて，$a=1,b=-20$

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2つの解が$4$と$-5$より，もとの2次方程式は，$(x-4)(x+5)=0$

したがって，$x^2+x-20=0$

よって，$x^2+ax+b=0$と係数を比較すると，$a=1,b=-20$

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(2)

$x=a$が$x^2+ax-8=0$の解であるから，$a^2+a\cdot a-8=0$

すなわち，$2a^2=8$より，$a=\pm 2$

(i) $a=2$のとき

方程式は，$x^2+2x-8=0$

したがって，$(x-2)(x+4)=0$

ゆえに，$x=2,-4$

したがって，他の解は，$x=-4$

(ii) $a=-2$のとき

方程式は，$x^2-2x-8=0$

したがって，$(x-4)(x+2)=0$

ゆえに，$x=4,-2$

したがって，他の解は，$x=4$よって，(i)，（ii）より，

\begin{align*}
&a=2\text{のとき，}&&\text{他の解は，}x=-4\\
&a=-2\text{のとき，}&&\text{他の解は，}x=4
\end{align*}

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% 例題I3.3.4の側注：2次方程式の係数決定（One More）★★
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% 側注1
数学IIで学習する，解と係数の関係を用いても求められる．

% 側注2
$\alpha,\beta$を解にもつ2次方程式は，$a(x-\alpha)(x-\beta)=0$と表される．与えられた方程式の$x^2$の係数が1であることから，$a=1$である．

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% 問題I3.3.4の側注：2次方程式の係数決定（One More）★★
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% 側注1
数学IIで学習する，解と係数の関係を用いても求められる．

% 側注2
$\alpha,\beta$を解にもつ2次方程式は，$a(x-\alpha)(x-\beta)=0$と表される．与えられた方程式の$x^2$の係数が1であることから，$a=1$である．

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% 例題I3.3.4の考え方・補足など：2次方程式の係数決定（One More）★★
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% 考え方
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2次方程式$ax^2+bx+c=0$の解が$\alpha$であるとき，$x=\alpha$を代入すると，$a\alpha^2+b\alpha+c=0$が成り立つ．

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% 余談
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2次方程式$ax^2+bx+c=0$の2つの解を$\alpha,\beta$とする．このとき，解と係数の関係$\alpha+\beta=-\frac{b}{a},\alpha\beta=\frac{c}{a}$が成り立つ．解と係数の関係を用いると，(1)は，$3+(-2)=-a,3\cdot(-2)=b$となり，$a=-1,b=-6$と簡単に$a,b$の値を求められる．