Solve by completing the square.
-$$y ^ { 2 } + 8 y - 5 = 0$$
A) $4 + \sqrt { 21 }$
B) $- 1 \pm \sqrt { 21 }$
C) $- 4 \pm \sqrt { 21 }$
D) $- 4 \pm 2 \sqrt { 21 }$

Question

Quizplus · Accepted Answer

To solve the equation $y^2 + 8y - 5 = 0$ by completing the square, first move the constant term to the other side: $y^2 + 8y = 5$. Then, add $(\frac{8}{2})^2 = 16$ to both sides to complete the square: $y^2 + 8y + 16 = 21$. This gives $(y + 4)^2 = 21$. Taking the square root of both sides gives $y + 4 = \pm \sqrt{21}$, so $y = -4 \pm \sqrt{21}$.

Solve by Completing the Square y2+8y−5=0y ^ { 2 } + 8 y - 5 = 0y2+8y−5=0

Solve by Completing the Square $y ^ { 2 } + 8 y - 5 = 0$