Question
Solve the following quadratic equations by factorization:
$3 x^2=-11 x-10$
$3 x^2=-11 x-10$
✓
Answer
$3 x^2=-11 x-10$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$\begin{cases}\because3\times10=30\\\therefore30=5\times6\\11=5+6\end{cases}$
$\Rightarrow 3 x^2+6 x+5 x+10=0$
$⇒ 3x(x + 2) + 5(x + 2) = 0$
$⇒ (x + 2)(3x + 5) = 0$
Either $x + 2 = 0$, then $x = -2$
Or $3x + 5 = 0$. then $3x = -5$
$\Rightarrow\text{x}=\frac{-5}{3}$
$\therefore$ Roots are $x = -2, \frac{-5}{3}$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$\begin{cases}\because3\times10=30\\\therefore30=5\times6\\11=5+6\end{cases}$
$\Rightarrow 3 x^2+6 x+5 x+10=0$
$⇒ 3x(x + 2) + 5(x + 2) = 0$
$⇒ (x + 2)(3x + 5) = 0$
Either $x + 2 = 0$, then $x = -2$
Or $3x + 5 = 0$. then $3x = -5$
$\Rightarrow\text{x}=\frac{-5}{3}$
$\therefore$ Roots are $x = -2, \frac{-5}{3}$
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