Solve the equation by completing the square. $4 x ^ { 2 } - 5 x + 28 = 0$
A) $x = \frac { 5 } { 8 } \pm \frac { \sqrt { 423 } i } { 8 }$
B) $x = \frac { 25 } { 8 } \pm \frac { \sqrt { 62 } i } { 2 }$
C) $x = \frac { 5 } { 8 } \pm \frac { \sqrt { 473 } } { 8 }$
D) $x = \frac { 5 } { 8 } \pm \frac { \sqrt { 62 } } { 8 }$
E) $x = \frac { 25 } { 8 } \pm \frac { \sqrt { 423 } } { 4 }$

Question

Quizplus · Accepted Answer

To solve $4x^2 - 5x + 28 = 0$ by completing the square, first divide the equation by 4 to simplify: $x^2 - \frac{5}{4}x + 7 = 0$. Then, find the value to complete the square: $\left(\frac{-5}{8}\right)^2 = \frac{25}{64}$. Adjust the equation to form a perfect square trinomial: $x^2 - \frac{5}{4}x + \frac{25}{64} = -7 + \frac{25}{64}$. Convert to vertex form: $\left(x - \frac{5}{8}\right)^2 = -\frac{423}{64}$. Solve for x: $x = \frac{5}{8} \pm \frac{\sqrt{423}i}{8}$, indicating complex solutions due to the negative value under the square root.

Solve the Equation by Completing the Square 4x2−5x+28=04 x ^ { 2 } - 5 x + 28 = 04x2−5x+28=0

Solve the Equation by Completing the Square $4 x ^ { 2 } - 5 x + 28 = 0$