Question

The roots of the equation \[{{x}^{3}}-6x+9=0\]?
1. -6
2. -9
3. 6
4. -3

Answer 1

Hint: In this problem, we have to find the roots of the given equation. Here we have a cubic equation, so we can first take a common term from the first two terms to get a factor which will be the factor of the equation. We can then use the quadratic formula to find the roots of the resulting quadratic equation.

Complete step by step answer:
Here we have to find the roots of the given equation,
\[{{x}^{3}}-6x+9=0\]
We can see that the given equation is a cubic equation, we can now write it as
\[\Rightarrow \left( x+3 \right)\left( {{x}^{2}}-3x+3 \right)=0\]
We can now substitute x = -3 in the given equation, we get
\[\begin{align}
  & \Rightarrow {{\left( -3 \right)}^{3}}-6\left( -3 \right)+9=0 \\
 & \Rightarrow -27+27=0 \\
\end{align}\]
Therefore, x = -3 is one of the roots of the given equation.
We can now find the roots of the equation \[{{x}^{3}}-3x+3=0\] using quadratic formulas, as
a = 1, b = -3 and c = 3
\[\begin{align}
  & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 3 \right)}}{2} \\
\end{align}\]
We can now simplify the above step, we get
\[\Rightarrow x=\dfrac{3\pm \sqrt{9-12}}{2}=\dfrac{3\pm \sqrt{-3}}{2}\]
We can see that the term inside the root is negative, so we can write it as,
\[\Rightarrow x=\dfrac{3\pm i\sqrt{3}}{2}\]
We can see that we have imaginary roots but we have only real roots in the option.

So, the correct answer is “Option 4”.

Note: We should always remember that if we have a negative term inside a square root, then it is an imaginary root, where \[\sqrt{-1}=i\]. In this problem, we have a real root and two imaginary roots. We should always remember the quadratic formula and substitute the required value to get the roots of the equation.