Solve by completing the square.
-$$\mathrm { y } ^ { 2 } + 4 \mathrm { y } - 3 = 0$$
A) $- 1 \pm \sqrt { 7 }$
B) $- 2 \pm 2 \sqrt { 7 }$
C) $- 2 \pm \sqrt { 7 }$
D) $2 + \sqrt { 7 }$

Question

Quizplus · Accepted Answer

To solve the equation $y^2 + 4y - 3 = 0$ by completing the square, first move the constant term to the other side: $y^2 + 4y = 3$. Then, add the square of half the coefficient of $y$ to both sides to complete the square: $(4/2)^2 = 4$. So, $y^2 + 4y + 4 = 3 + 4$, which simplifies to $(y + 2)^2 = 7$. Taking the square root of both sides gives $y + 2 = \pm \sqrt{7}$, so $y = -2 \pm \sqrt{7}$.

Solve by Completing the Square y2+4y−3=0\mathrm { y } ^ { 2 } + 4 \mathrm { y } - 3 = 0y2+4y−3=0

Solve by Completing the Square $\mathrm { y } ^ { 2 } + 4 \mathrm { y } - 3 = 0$