Determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then
write and factor the trinomial.
-$$x ^ { 2 } - 12 x - 5 = 0$$
A) $\{ 6 - \sqrt { 41 } , 6 + \sqrt { 41 } \}$
B) $\{ 6 - \sqrt { 5 } , 6 + \sqrt { 5 } \}$
C) $\{ - 6 - \sqrt { 41 } , - 6 + \sqrt { 41 } \}$
D) $\{ 12 - \sqrt { 149 } , 12 + \sqrt { 149 } \}$

Question

Quizplus · Accepted Answer

To make a perfect square trinomial, the constant term should be the square of half the coefficient of the linear term. In this case, the coefficient of the linear term is -12, so half of that is -6. The square of -6 is 36, so we need to add 36 to the binomial to make it a perfect square trinomial. The resulting trinomial is $x^2 - 12x + 36$, which can be factored as $(x - 6)^2$.

Determine the Constant That Should Be Added to the Binomial x2−12x−5=0x ^ { 2 } - 12 x - 5 = 0x2−12x−5=0

Determine the Constant That Should Be Added to the Binomial $x ^ { 2 } - 12 x - 5 = 0$