College Algebra Question 2(x/x+2)^2 -5(x/x+2)-3=0?

Presuming that is supposed to be:

2[ x / (x + 2)]² - 5[ x / (x + 2)] - 3 = 0

The first thing to notice is that if x = -2, you are dividing by zero, so that cannot be a solution.

The next thing to notice is that your squared expression matches that in the second term, so we can do a temporary substitution. Let's do this:

u = x / (x + 2)

After that substitution, you have a more straightforward quadratic:

2u² - 5u - 3 = 0

This factors:

(2u + 1)(u - 3) = 0

So we have two possible values for u:

u = -1/2 and 3

Now let's substitute our expression with "x" in it back in for u to have two equations which we can solve for x:

x / (x + 2) = -1/2 and x / (x + 2) = 3

Let's start with multiplying both sides by (x + 2):

x = -(x + 2) / 2 and x = 3(x + 2)

The first equation, let's multiply both sides by -2, then distribute the 3 in the second:

-2x = x + 2 and x = 3x + 6

Let's subtract x from both sides in the first equation and 3x from both sides in the second:

-3x = 2 and -2x = 6

And finally, divide both sides by -3 and -2 respectively:

x = -2/3 and x = -3

Neither of these solutions is the -2 we are looking for, so you have two solutions to your equation.

Let x / (x + 2) = u

2u^2 - 5u - 3 = 0

That looks nicer

u = (5 +/- sqrt(25 + 24)) / 4
u = (5 +/- 7) / 4
u = 12/4 , -2/4
u = 3 , -1/2
x/(x + 2) = 3 , -1/2

x / (x + 2) = 3
x = 3 * (x + 2)
x = 3x + 6
x - 3x = 6
-2x = 6
x = -6/2
x = -3

x / (x + 2) = -1/2
2 * x = -1 * (x + 2)
2x = -x - 2
2x + x = -2
3x = -2
x = -2/3

x = -3 , -2/3

Test them both.

2(x/x+2)^2 -5(x/x+2)-3=0
if I took that as you typed it, it would be
2(1+2)^2 -5(1+2)-3=0
but I don't think that is what you intended

2(x/(x+2))² - 5(x/(x+2)) - 3 = 0
multiply by (x+2)²
2x² – 5x(x+2) – 3(x+2)² = 0
2x² – 5x² – 10x – 3(x² + 4x + 4) = 0
– 3x² – 10x – 3x² – 12x – 12 = 0
–6x² – 22x – 12 = 0
3x² + 11x + 6 = 0
use a quadratic solver or the quadratic equation.
x = –2/3, –3

x=-3
x=-2/3
for
2(x/(x+2))² -5(x/(x+2))-3=0
with u=(x/(x+2))
2u²-5u-3=0