PERSAMAAN KUADRAT BARU
Persamaan kuadrat yang memiliki akar-akar \(\alpha\) dan \(\beta\) dapat dibentuk dengan cara:
Cara 1
\((x - \alpha)(x - \beta) = 0\)
Cara 2
\(x^2 - (\alpha + \beta)x + \alpha \:.\: \beta = 0\)
Contoh 01
Tentukan persamaan kuadrat yang akar-akarnya 5 dan −3
\(\alpha = 5\) dan \(\beta = -3\)
Cara 1
\begin{equation*} \begin{split} & (x - \alpha)(x - \beta) = 0 \\\\ & (x - 5)(x + 3) = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x ^2 - 2x - 15 = 0} \end{split} \end{equation*}
Cara 2
\begin{equation*} \begin{split} & x^2 - (\alpha + \beta)x + \alpha \:.\: \beta = 0 \\\\ & x^2 - (5 - 3)x + 5 \:.\: -3 = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x ^2 - 2x - 15 = 0} \end{split} \end{equation*}
Contoh 02
Akar-akar persamaan kuadrat \(2x^2 + 3x - 7 = 0\) adalah \(x_1\) dan \(x_2\).
Susunlah persamaan kuadrat baru yang akar-akarnya \((x_1 - 1)\) dan \((x_2 - 1)\)
\(\alpha = x_1 - 1\) dan \(\beta = x_2 - 1\)
\(2x^2 + 3x - 7 = 0\)
\begin{equation*} \begin{split} & x_1 + x_2 = - \frac ba \\\\ & \bbox[5px, border: 2px solid blue] {x_1 + x_2 = - \frac 32} \end{split} \end{equation*}
\begin{equation*} \begin{split} & x_1 \:.\: x_2 = \frac ca \\\\ & \bbox[5px, border: 2px solid blue] {x_1 \:.\: x_2 = \frac {-7}{2}} \end{split} \end{equation*}
Menentukan \(\alpha + \beta\)
\begin{equation*} \begin{split} & \alpha + \beta = x_1 - 1 + x_2 - 1 \\\\ & \alpha + \beta = x_1 + x_2 - 2 \\\\ & \alpha + \beta = -\frac{3}{2} - 2 \\\\ & \bbox[5px, border: 2px solid blue] {\alpha + \beta = -\frac{7}{2}} \end{split} \end{equation*}
Menentukan \(\alpha \:.\: \beta\)
\begin{equation*} \begin{split} & \alpha \:.\: \beta = (x_1 - 1)(x_2 - 1) \\\\ & \alpha \:.\: \beta = x_1 \:.\: x_2 - x_1 - x_2 + 1\\\\ & \alpha \:.\: \beta = x_1 \:.\: x_2 - (x_1 + x_2) + 1\\\\ & \alpha \:.\: \beta = \frac{-7}{2} + \frac{3}{2} + 1\\\\ & \bbox[5px, border: 2px solid blue] {\alpha \:.\: \beta = -1} \end{split} \end{equation*}
Persamaan kuadrat baru
\begin{equation*} \begin{split} & x^2 - (\alpha + \beta)x + \alpha \:.\: \beta = 0 \\\\ & x^2 - \left(-\frac{7}{2} \right) x - 1 = 0 \\\\ & x^2 + \frac{7}{2} x - 1 = 0 \quad {\color {blue} \dotso \: \times 2} \\\\ & \bbox[5px, border: 2px solid magenta] {2x^2 + 7x - 2 = 0} \end{split} \end{equation*}
SOAL LATIHAN