Question

Determine which equation has the same solutions as the given equation
2x^2 - 12x - 50 = 0
A. (x-3)^2 = 16
B. (x - 6)^2 = 19
C. (x - 6)^2 = 31
D. (x - 3)^2 = 34

Answer 1

Step-by-step explanation:

here is the answer ....feel free to ask if u don't get it

Answer 2

Answer:

D. $(x-3)^2=34$

Step-by-step explanation:

Its looks like they have complete the square for the given equation.

So let's do that.

$2x^2-12x-50=0$

I see all of my terms are even so I can easily divide each of them by 2.

I'm going to divide both sides by 2:

$x^2-6x-25=0$

The first step in completing the square is to make sure the coefficient of $x^2$ is 1 so we can use this easy formula:

$x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$.

So I see in place of k I have -6.

I'm going to add $(\frac{-6}{2})^2$ on both sides.

$x^2-6x+(\frac{-6}{2})^2-25=(\frac{-6}{2})^2$

See these first three terms here can be written using the mentioned formula:

$(x+\frac{-6}{2})^2-25=(-\frac{-6}{2})^2$

Let's simplify a bit:

$(x+(-3))^2-25=(-3)^2$

$(x-3)^2-25=9$

Add 25 on both sides:

$(x-3)^2=9+25$

$(x-3)^2=34$