Solve the equation by completing the square.
-$$4 \mathrm { x } ^ { 2 } + 1 = 3 \mathrm { x }$$
A) $\left\{ \frac { 3 - i \sqrt { 7 } } { 8 } , \frac { - 3 + i \sqrt { 7 } } { 8 } \right\}$
B) $\left\{ \frac { 3 - i \sqrt { 7 } } { 8 } , \frac { 3 + i \sqrt { 7 } } { 8 } \right\}$
C) $\left\{ \frac { - 3 - \mathrm { i } \sqrt { 7 } } { 8 } , \frac { - 3 + \mathrm { i } \sqrt { 7 } } { 8 } \right\}$
D) $\left\{ \frac { - 3 - \mathrm { i } \sqrt { 7 } } { 8 } , \frac { 3 + \mathrm { i } \sqrt { 7 } } { 8 } \right\}$

Question

Quizplus · Accepted Answer

To solve by completing the square, rewrite the equation as $4x^2 - 3x + 1 = 0$. Divide by 4: $x^2 - \frac{3}{4}x = -\frac{1}{4}$. Add $(\frac{3}{8})^2 = \frac{9}{64}$ to both sides to complete the square: $(x - \frac{3}{8})^2 = \frac{-7}{64}$. Solve for $x$: $x = \frac{3}{8} \pm \frac{i\sqrt{7}}{8}$.

Solve the Equation by Completing the Square 4x2+1=3x4 \mathrm { x } ^ { 2 } + 1 = 3 \mathrm { x }4x2+1=3x

Solve the Equation by Completing the Square $4 \mathrm { x } ^ { 2 } + 1 = 3 \mathrm { x }$