Question

Find all solutions to the equation. -\dfrac{2}{x(x+2)}=\dfrac{x+3}{x+2}− x(x+2) 2 ​ = x+2 x+3 ​ minus, start fraction, 2, divided by, x, (, x, plus, 2, ), end fraction, equals, start fraction, x, plus, 3, divided by, x, plus, 2, end fraction Choose all answers that apply: Choose all answers that apply:

Answer 1

Answer:

$x=-1$

Step-by-step explanation:

We have been given an equation $\frac{-2}{x(x+2)}=\frac{x+3}{x+2}$. We are asked to find the solutions of our given equation.

First of all, we will cross multiply our given equation as:

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

Subtract $x(x+2)(x+3)$ from both sides:

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

$-2(x+2)-x(x+2)(x+3)=0$

Now we will factor out $(x+2)$.

$(x+2)(-2-x(x+3))=0$

$(x+2)(-2-x^2-3x)=0$

$(x+2)(-(x^2+3x+2))=0$

Now we will factor by splitting the middle term.

$-(x+2)(x^2+2x+x+2)=0$

$-(x+2)(x+2)(x+1)=0$

$-(x+2)^2(x+1)=0$

Now we will use zero product property and solve for x as:

$-(x+2)^2=0,(x+1)=0$

$(x+2)=0,(x+1)=0$

$x=-2,x=-1$

Now, we will check for extraneous solutions.

We can see that $x=-2$ will make both denominators zero because our function is not defined at  $x=-2$.

Therefore, the solution for our given equation is $x=-1$.

Answer 2

Answer:

zk

Step-by-step explanation:

zc