Trik Jitu #7 - Persamaan Kuadrat

$latex Jika\ persamaan\ kuadrat\ mempunyai\ akar-akar\ nyata\ (real)\ dan\ sama,\ maka\ diskriminannya\ nol.\\\\Jumlah\ kedua\ akar\ persamaan\ kuadrat\ ax^2+bx+c=0\ ditentukan\ oleh\ {\color{red}x_{1}+x_{2}=-\frac{b}{a}}.$

Contoh:

$latex Persamaan\ kuadrat\ (k+2)x^2-(2k-1)x+k-1=0\ akar-akarnya\ nyata\ dan\ sama.\\Jumlah\ kedua\ akar\ persamaan\ tersebut\ adalah\ ....\\\\A.\ \ \frac{9}{8}\\\\B.\ \ \frac{8}{9}\\\\C.\ \ \frac{5}{2}\\\\D.\ \ \frac{2}{5}\\\\E.\ \ \frac{1}{5}$

Jawaban: D
$latex \begin{array}{rcl}D&=&0\\b^2-4ac&=&0\\\begin{bmatrix}-(2k-1)\end{bmatrix}^2-4(k+2)(k-1)&=&0\\(2k-1)^2-4(k^2+k-2)&=&0\\4k^2-4k+1-4k^2-4k+8&=&0\\-8k+9&=&0\\8k&=&9\\k&=&\frac{9}{8}\end{array}$

$latex \begin{array}{rcl}x_{1}+x_{2}&=&\large -\frac{b}{a}\\\\&=&-\frac{-(2k-1)}{k+2}\\\\&=&\frac{2k-1}{k+2}\\\\&=&\frac{2\big(\frac{9}{8}\big)-1}{\frac{9}{8}+2}\quad ........\ dikalikan\ \frac{8}{8}\\\\&=&\frac{18-8}{9+16}\\\\&=&\frac{10}{25}\\\\&=&\frac{2}{5}\end{array}$

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