Question

The real root of the equation x³ - 6x + 9 = 0 is
1. -6
2. -9
3. 6
4. -3

Answer 1

Correct Answer - Option 4 : -3

Concept:

Conditions for real roots: Discriminant ≥ 0 or b² - 4ac ≥ 0

Conditions for imaginary roots: b2 - 4ac < 0

Calculation:

The equation is x3 - 6x + 9 = 0

Let x = -3

⇒ (-3)³ - 6(-3) + 9 = 0

⇒ -27 + 18 + 9 = 0

Hence (x + 3) is a factor of x3 - 6x + 9 = 0

Now the equation can be written as

(x + 3)(x² - 3x + 3)

One root will be x + 3 = 0, x = - 3

Consider (x2 - 3x + 3)

b² - 4ac = 9 - 4(1)(3) = - 3

So, b2 - 4ac < 0

Hence, The real root of the equation x3 - 6x + 9 = 0 is - 3