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

Question

The real root of the equation x3-6x+9 = 0 is (A) -6 (B) -9 (C) 6 (D) -3

Tardigrade Admin · Accepted Answer

Given, x3-6 x+9=0 ⇒ (x+3)(x2-3 x+3)=0 ⇒ x=-3 or x2-3 x+3=0 Now, Discriminant, D=√9-4 × 3 =√-3 imaginary Hence, real roots of the given equation is x=-3

Q. The real root of the equation $x^{3} - 6 x + 9 = 0$ is

-6

-9

6

-3

Given, $x^{3} - 6 x + 9 = 0$
$\Rightarrow (x + 3) (x^{2} - 3 x + 3) = 0$
$\Rightarrow x = - 3$ or $x^{2} - 3 x + 3 = 0$
Now, Discriminant, $D = 9 - 4 \times 3$
$= - 3$ imaginary
Hence, real roots of the given equation is
$x = - 3$

Q. The real root of the equation x3−6x+9=0 is

-6

-9

6

-3

Given, x3−6x+9=0 ⇒(x+3)(x2−3x+3)=0 ⇒x=−3 or x2−3x+3=0 Now, Discriminant, D=9−4×3​ =−3​ imaginary Hence, real roots of the given equation is x=−3

Q. The real root of the equation $x^{3} - 6 x + 9 = 0$ is

Given, $x^{3} - 6 x + 9 = 0$
$\Rightarrow (x + 3) (x^{2} - 3 x + 3) = 0$
$\Rightarrow x = - 3$ or $x^{2} - 3 x + 3 = 0$
Now, Discriminant, $D = 9 - 4 \times 3$
$= - 3$ imaginary
Hence, real roots of the given equation is
$x = - 3$