Solve by completing the square.
-$$y ^ { 2 } + 3 y - 9 = 0$$
A) $\frac { - 3 - 3 \sqrt { 5 } } { 2 }$
B) $- 3 \pm 3 \sqrt { 5 }$
C) $\frac { - 3 \pm 3 \sqrt { 5 } } { 2 }$
D) $\frac { 3 + 3 \sqrt { 5 } } { 2 }$

Question

Quizplus · Accepted Answer

To solve the equation $y^2 + 3y - 9 = 0$ by completing the square, first move the constant term to the other side: $y^2 + 3y = 9$. Then, add $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ to both sides to complete the square: $y^2 + 3y + \frac{9}{4} = 9 + \frac{9}{4} = \frac{36}{4} + \frac{9}{4} = \frac{45}{4}$. This gives $\left(y + \frac{3}{2}\right)^2 = \frac{45}{4}$. Taking the square root of both sides gives $y + \frac{3}{2} = \pm \sqrt{\frac{45}{4}}$, which simplifies to $y + \frac{3}{2} = \pm \frac{3\sqrt{5}}{2}$. Solving for $y$ yields $y = -\frac{3}{2} \pm \frac{3\sqrt{5}}{2}$, which simplifies to $y = \frac{-3 \pm 3\sqrt{5}}{2}$.

Solve by Completing the Square y2+3y−9=0y ^ { 2 } + 3 y - 9 = 0y2+3y−9=0

Solve by Completing the Square $y ^ { 2 } + 3 y - 9 = 0$