Trik Jitu #6 - Persamaan Kuadrat

$latex Jika\ persamaan\ kuadrat\ mempunyai\ akar-akar\ nyata\ (real)\ maka\ diskriminannya\ positif\ atau\ nol.\\\\{\bf{Catatan:}}\\Diskriminan\ persamaan\ kuadrat\ ax^2+bx+c=0\ ditentukan\ oleh\ {\color{red}D=b^2-4ac}.$

Contoh:

$latex Persamaan\ kuadrat\ x^2+(m-2)x+9=0\ akar-akarnya\ nyata.\ Nilai\ m\ yang\ memenuhi\ adalah\ ....\\A.\ \ m\le -4\ atau\ m\ge 8\\B.\ \ m\le -8\ atau\ m\ge 4\\C.\ \ m\le -4\ atau\ m\ge 10\\D.\ \ -4\le x\le 8\\E.\ \ -8\le x\le 4$

Jawaban: A

$latex \begin{array}{rcl}D&\ge &0\\b^2-4ac&\ge &0\\(m-2)^2-4.1.9&\ge &0\\(m-2)^2-6^2&\ge &0\\(m-2+6)(m-2-6)&\ge &0\ \ ........\ {\color{blue}(1)}\\(m+4)(m-8)&\ge &0\\m\ \le \ -4\ \ atau\ \ m&\ge &8\ \ ........\ {\color{blue}(2)}\end{array}$

$latex {\bf{Keterangan:}}\\{\color{blue}(1)\ \ Faktorisasi}:\ a^2-b^2=(a+b)(a-b)\\{\color{blue}(2)\ \ Penyelesaian}:\ Bentuk\ a(x-x_{1})(x-x_{2})\ dengan\ x_{1}<x_{2}\\\begin{matrix} \ \ \quad \end{matrix}a(x-x_{1})(x-x_{2})=0\qquad \qquad \qquad \ \ \Rightarrow \qquad x=x_{1}\ dan\ x=x_{2}\\\begin{matrix} \ \ \quad \end{matrix}a(x-x_{1})(x-x_{2})>0\ dan\ a>0\qquad \Rightarrow \qquad x<x_{1}\ atau\ x>x_{2}\\\begin{matrix} \ \ \quad \end{matrix}a(x-x_{1})(x-x_{2})<0\ dan\ a>0\qquad \Rightarrow \qquad x_{1}<x<x_{2}.$

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