Question

I could really use the help! problem is down below in the image. :)

not in a rush, but I could use the answer soon.

Answer 1

There is no extraneous solution for this one. x=-2 and x=-9

Answer 2

Answer:

The two posible solutions are x = -9  and  x = -2

Step-by-step explanation:

We have the equation:

$\sqrt{7-x}$ = x+5

As shown in the figure we can express this equation in this form:

($\sqrt{7-x}$)^2 = (x+5)^2

Solving:

7-x = x^2 + 10x +25

We want to have the equation equal to 0:

0 = x^2 + 11x + 18 (this is already in the picture)

Now we are going to factor the equation to get 2 solutions for x, (in this case we will have 2 solutions for x because it is a quadratic equation).

To factor the equation we have to find two numbers that when you multiply them you get 18 and when you add them you get 11

These two numbers will be 2 and 9; if you multiply 2*9=18 and 2+9=11.

Now we factor:

(x+9)(x+2)=0

and we can work as separate equations:

x+9=0

x+2=0

Solving:

x = -9

x = -2

We can prove this by substituting the values of x in the first equation $\sqrt{7-x}$ = x+5:

$\sqrt{7-(-9)}$ = -9+5

$\sqrt{16}$ = -4

-4 = -4

$\sqrt{7-(-2)}$ = -2+5

$\sqrt{9}$  = 3

3 = 3

In this case, we don´t have extraneous solutions, both are true solutions for this equation.