Solve by completing the square.
-$$3 x ^ { 2 } - 7 x - 5 = x ^ { 2 } + 2 x - x ( x + 1 ) + 1$$
A) $\frac { 4 \pm \sqrt { 34 } } { 3 }$
B) $\frac { 6 \pm \sqrt { 34 } } { 3 }$
C) $\frac { 5 \pm \sqrt { 23 } } { 3 }$
D) $\frac { 4 \pm \sqrt { 31 } } { 3 }$

Question

Quizplus · Accepted Answer

First, simplify the equation: $3x^2 - 7x - 5 = x^2 + 2x - x^2 - x + 1$, which simplifies to $3x^2 - 7x - 5 = x + 1$. Rearrange to get $3x^2 - 8x - 6 = 0$. Divide by 3 to get $x^2 - \frac{8}{3}x - 2 = 0$. Completing the square involves making the left side of the equation a perfect square trinomial: $x^2 - \frac{8}{3}x + \left(\frac{8}{6}\right)^2 = 2 + \left(\frac{8}{6}\right)^2$, which simplifies to $x^2 - \frac{8}{3}x + \frac{16}{9} = \frac{18}{9} + \frac{16}{9}$. This gives $x^2 - \frac{8}{3}x + \frac{16}{9} = \frac{34}{9}$, so $(x - \frac{4}{3})^2 = \frac{34}{9}$. Taking the square root of both sides gives $x - \frac{4}{3} = \pm \sqrt{\frac{34}{9}}$, so $x = \frac{4}{3} \pm \frac{\sqrt{34}}{3}$.

Solve by Completing the Square 3x2−7x−5=x2+2x−x(x+1)+13 x ^ { 2 } - 7 x - 5 = x ^ { 2 } + 2 x - x ( x + 1 ) + 13x2−7x−5=x2+2x−x(x+1)+1

Solve by Completing the Square $3 x ^ { 2 } - 7 x - 5 = x ^ { 2 } + 2 x - x ( x + 1 ) + 1$