Solve the equation in the real number system.
-$$2 x ^ { 3 } - x ^ { 2 } - 6 x + 3 = 0$$
A) $\{ 2 , \sqrt { 3 } , - \sqrt { 3 } \}$
B) $\left\{ - \frac { 1 } { 2 } , \sqrt { 3 } , - \sqrt { 3 } \right\}$
C) $\left\{ \frac { 1 } { 2 } , \sqrt { 3 } , - \sqrt { 3 } \right\}$
D) $\{ - 2 , \sqrt { 3 } , - \sqrt { 3 } \}$

Question

Quizplus · Accepted Answer

The given equation $2x^3 - x^2 - 6x + 3 = 0$ can be factored into $(x-1/2)(2x^2-3)=0$, which gives the roots $x = 1/2$, $x = \sqrt{3}$, and $x = -\sqrt{3}$.

Solve the Equation in the Real Number System 2x3−x2−6x+3=02 x ^ { 3 } - x ^ { 2 } - 6 x + 3 = 02x3−x2−6x+3=0

Solve the Equation in the Real Number System $2 x ^ { 3 } - x ^ { 2 } - 6 x + 3 = 0$