Solve the equation by completing the square.
-$$5 x ^ { 2 } - 2 x - 1 = 0$$
A) $\left\{ \frac { 1 - \sqrt { 6 } } { 5 } , \frac { 1 + \sqrt { 6 } } { 5 } \right\}$
B) $\left\{ \frac { 5 - \sqrt { 6 } } { 25 } , \frac { 5 + \sqrt { 6 } } { 25 } \right\}$
C) $\left\{ - 1 , \frac { 7 } { 5 } \right\}$
D) $\left\{ \frac { - 1 - \sqrt { 6 } } { 5 } , \frac { - 1 + \sqrt { 6 } } { 5 } \right\}$

Question

Quizplus · Accepted Answer

To solve $5x^2 - 2x - 1 = 0$ by completing the square, first divide the equation by 5 to get $x^2 - \frac{2}{5}x - \frac{1}{5} = 0$. Then, move the constant term to the other side to get $x^2 - \frac{2}{5}x = \frac{1}{5}$. To complete the square, add $\left(\frac{-2}{2 \cdot 5}\right)^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}$ to both sides, resulting in $x^2 - \frac{2}{5}x + \frac{1}{25} = \frac{1}{5} + \frac{1}{25}$. This simplifies to $\left(x - \frac{1}{5}\right)^2 = \frac{6}{25}$. Taking the square root of both sides gives $x - \frac{1}{5} = \pm \sqrt{\frac{6}{25}}$, or $x = \frac{1}{5} \pm \frac{\sqrt{6}}{5}$, which simplifies to the solutions $\frac{1 - \sqrt{6}}{5}$ and $\frac{1 + \sqrt{6}}{5}$.

Solve the Equation by Completing the Square 5x2−2x−1=05 x ^ { 2 } - 2 x - 1 = 05x2−2x−1=0

Solve the Equation by Completing the Square $5 x ^ { 2 } - 2 x - 1 = 0$