Use the method of completing the square to solve the quadratic equation. $3 x ^ { 3 } + 7 x - 1 = 0$
A) $x = \frac { - 7 + \sqrt { 61 } } { 2 }$
B) $x = \frac { - 7 \pm \sqrt { 61 } } { 6 }$
C) $x = \frac { \pm 7 \pm \sqrt { 61 } } { 2 }$
D) $x = \frac { \pm 7 + \sqrt { 61 } } { 2 }$
E) $x = \frac { \pm 7 \pm \sqrt { 61 } } { 6 }$

Question

Quizplus · Accepted Answer

The given equation is $3x^2 + 7x - 1 = 0$, not $3x^3 + 7x - 1 = 0$ as stated. To solve it by completing the square, we first divide the entire equation by 3 to normalize the coefficient of $x^2$, getting $x^2 + \frac{7}{3}x - \frac{1}{3} = 0$. Completing the square involves rearranging the equation to the form $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$, where $a = 1$, $b = \frac{7}{3}$, and $c = -\frac{1}{3}$. Solving the equation after completing the square gives roots that match the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which simplifies to $x = \frac{-7 \pm \sqrt{61}}{6}$ for the given coefficients.

Use the Method of Completing the Square to Solve the Quadratic