Question

Solve each of the following quadratic equations:

$3x^2-2x-1=0$

Answer 1

We write, $-2x=-3x+x$ as $3x^2\times \left(-1\right)=-3x^2=\left(-3x\right)\times x$

$$\begin{gathered} \therefore 3x^2-2x-1=0 \\ \Rightarrow 3x^2-3x+x-1=0 \\ \Rightarrow 3x\left(x-1\right)+1\left(x-1\right)=0 \\ \Rightarrow \left(x-1\right)\left(3x+1\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-1=0\mathrm{or}3x+1=0 \\ \Rightarrow x=1\mathrm{or}x=-\frac{1}{3} \end{gathered}$$
Hence, the roots of the given equation are $1$ and $-\frac{1}{3}$.