Solve by completing the square.
-$$\mathrm { q } ^ { 2 } + 4 \mathrm { q } - 9 = 0$$
A) $2 + \sqrt { 13 }$
B) $- 2 \pm \sqrt { 13 }$
C) $- 1 \pm \sqrt { 13 }$
D) $- 2 \pm 2 \sqrt { 13 }$

Question

Quizplus · Accepted Answer

To solve the quadratic equation $q^2 + 4q - 9 = 0$ by completing the square, we first move the constant term to the other side: $q^2 + 4q = 9$. Then, we add $(\frac{4}{2})^2 = 4$ to both sides to complete the square: $q^2 + 4q + 4 = 13$, which simplifies to $(q + 2)^2 = 13$. Taking the square root of both sides gives $q + 2 = \pm \sqrt{13}$, so $q = -2 \pm \sqrt{13}$.

Solve by Completing the Square q2+4q−9=0\mathrm { q } ^ { 2 } + 4 \mathrm { q } - 9 = 0q2+4q−9=0

Solve by Completing the Square $\mathrm { q } ^ { 2 } + 4 \mathrm { q } - 9 = 0$