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The real root of the equation $x^{3}-6 x+9=0$ is
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Verified Answer
The correct answer is:
$-3$
Given, $x^{3}-6 x+9=0$
$\Rightarrow \quad(x+3)\left(x^{2}-3 x+3\right)=0$
$\Rightarrow \quad \mathrm{x}=-3 \quad$ or $\quad \mathrm{x}^{2}-3 \mathrm{x}+3=0$
Now, Discriminant, $\mathrm{D}=\sqrt{9-4 \times 3}$
$=\sqrt{-3}$ imaginary
Hence, real roots of the given equation is
$x=-3$
$\Rightarrow \quad(x+3)\left(x^{2}-3 x+3\right)=0$
$\Rightarrow \quad \mathrm{x}=-3 \quad$ or $\quad \mathrm{x}^{2}-3 \mathrm{x}+3=0$
Now, Discriminant, $\mathrm{D}=\sqrt{9-4 \times 3}$
$=\sqrt{-3}$ imaginary
Hence, real roots of the given equation is
$x=-3$
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