Solve $u ^ { 2 } - 7 u - 4 = 0$ by completing the square, if possible.
A) $y = \frac { \sqrt { 65 } } { 2 } , y = - \frac { \sqrt { 65 } } { 2 }$
B) $y = \frac { 7 } { 2 } + \frac { \sqrt { 65 } } { 2 } , y = \frac { 7 } { 2 } - \frac { \sqrt { 65 } } { 2 }$
C) $y = 7 + \frac { \sqrt { 65 } } { 2 } , y = 7 - \frac { \sqrt { 65 } } { 2 }$
D) $y = - \frac { 7 } { 2 } + \frac { \sqrt { 65 } } { 2 } , y = - \frac { 7 } { 2 } - \frac { \sqrt { 65 } } { 2 }$
E)no solutions

Question

Quizplus · Accepted Answer

To solve $u^2 - 7u - 4 = 0$ by completing the square, first move the constant term to the other side: $u^2 - 7u = 4$. Then, add $\left(\frac{-7}{2}\right)^2 = \frac{49}{4}$ to both sides to complete the square: $u^2 - 7u + \frac{49}{4} = 4 + \frac{49}{4}$, which simplifies to $u^2 - 7u + \frac{49}{4} = \frac{65}{4}$. This gives $\left(u - \frac{7}{2}\right)^2 = \frac{65}{4}$. Taking the square root of both sides gives $u - \frac{7}{2} = \pm \frac{\sqrt{65}}{2}$, leading to $u = \frac{7}{2} \pm \frac{\sqrt{65}}{2}$.

Solve u2−7u−4=0u ^ { 2 } - 7 u - 4 = 0u2−7u−4=0

Solve $u ^ { 2 } - 7 u - 4 = 0$