KCET · Maths · Quadratic Equation
The real root of the equation \(x^{3}-6 x+9=0\) is
- A \(-6\)
- B \(-9\)
- C 6
- D \(-3\)
Answer & Solution
Correct Answer
(D) \(-3\)
Step-by-step Solution
Detailed explanation
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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