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The roots of the equation $x^3-3 x-2=0$ are
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2659 Upvotes
Verified Answer
The correct answer is:
$-1,-1,2$
Given equation is
$x^3-3 x-2=0$
Now, put $x=-1$, we get
$\begin{aligned}
(-1)^3-3(-1)-2 & =0 \\
\Rightarrow \quad 0 & =0
\end{aligned}$
Given equation can be written as
$\begin{aligned}
& \therefore \quad(x+1)\left(x^2-x-2\right)=0 \\
& \Rightarrow \quad(x+1)(x+1)(x-2)=0 \\
& \Rightarrow \quad x=-1,-1,2 \\
&
\end{aligned}$
$x^3-3 x-2=0$
Now, put $x=-1$, we get
$\begin{aligned}
(-1)^3-3(-1)-2 & =0 \\
\Rightarrow \quad 0 & =0
\end{aligned}$
Given equation can be written as
$\begin{aligned}
& \therefore \quad(x+1)\left(x^2-x-2\right)=0 \\
& \Rightarrow \quad(x+1)(x+1)(x-2)=0 \\
& \Rightarrow \quad x=-1,-1,2 \\
&
\end{aligned}$
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