Question

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

Answer 1

Answer:

x=-2 (the actual solution)

x=-5 (the extraneous solution)

Step-by-step explanation:

First step is to isolate the radical part (the part with the square root).

So we will add 4 on both sides first:

$\sqrt{x+6}=x+4$

To get rid of square we are going to square both sides:

$x+6=(x+4)^2$

We are going to expand right hand side (you can use "F.O.I.L":

$x+6=x^2+4x+4x+16$

Subtract x on both sides and subtract 6 on both sides:

$0=x^2+4x+4x-x+16-6$

Combine like terms:

$0=x^2+7x+10$

We will now factor:

$0=(x+2)(x+5)$ (Since 2(5)=10 and 2+5=7)

Set both factors equal to 0:

x+2=0 or x+5=0

Solve first equation by subtracting 2 on both sides to obtain x=-2.

Solve second equation by subtracting 5 on both sides to obtain x=-5.

Now let's check to see if these are both actually indeed solutions to the equation we began with:

$\sqrt{x+6}-4=x$ for $x=-2$:

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$ is a true equation so $x=-2$ is indeed a solution.

$\sqrt{x+6}-4=x$ for $x=-5$:

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3=-5$ is false so $x=-5$ is what we call the extraneous solution.

Answer 2

Answer:

-5 [Extraneous Solution], -2 = x

Step-by-step explanation:

1. √[x + 6] = x + 4 [Move 4 to the right side of the equivalence symbol]
2. x + 6 = x² + 8x + 16 [Square both sides to get rid of the radical, leaving you to FOIL the right side of the equivalence symbol]
3. -7x - 10 = x² [Move -8x - 10 to the left side of the equivalence symbol, leaving you with x² on the left]
4. x² + 7x + 10 [Rewrite in Quadratic Standard Form]
5. [2 + x][5 + x] → Factor by finding two numbers that when added up to 7, they also multiply up to 10.
6. -2, -5 = x [Set each expression equal to zero and solve for x.]
7. Plug these x-coordinates back into the given equation to search for the extraneous solution, which is a root that results in a false statement.
8. √[-2 + 6] - 4 = -2 → √4 = ±2 [In this case, it is NON-NEGATIVE] → 2 - 4 = -2 >> -2 ☑; √[-5 + 6] - 4 = -5 → √1 = ±1 [Although the negative works, we still want the NON-NEGATIVE] → 1 - 4 ≠ -5 >> -5 ☒
9. We come to the conclusion that -5 = x is the extraneous solution.

I am joyous to assist you anytime.

**Whenever we are dealing with extraneous solutions, we always want the NON-NEGATIVE roots.