Solve Radical Equations with Extraneous Solutions Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous. Thank You!
1. √x+1+5 = x

2. √5x-x = 0

3. √2x+1+1 =x

4. √3x-2+3 = 4x

5. 2√4-2x = 2-x

Question

Solve Radical Equations with Extraneous Solutions Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous. Thank You!
1. √x+1+5 = x

2. √5x-x = 0

3. √2x+1+1 =x

4. √3x-2+3 = 4x

5. 2√4-2x = 2-x

anonymous · Answer

hmmm i cant help you sorry :c

snowsurf · Answer

I will do one for you. Let's do the first one.
√x+1+5 = x

Solve it like a algebra equation. Combine like terms and isolate x.

$$\sqrt{x} + 6 = x$$
$$\sqrt{x}=x-6$$

Now take square roots of both sides 

$$x=(x-6)^{2}$$
Now use FOIL 

x = $$x = x ^{2}-12x+36$$

$$x ^{2}-13x+36=0$$

Factor this equation to find the roots.

(x-4)(x-9) = 0
So the solutions are x = 9 and x =4

But plug in x=4 and x=9 into the equation
√9+1+5 = 9

3 + 1 + 5 = 9
9 = 9

Therefore 9 is a solution to the square root equation.

√4+1+5 = 4

2+1+5=4
8=4
4 is not a solution to the square root equation but its a solution to the quadratic equation. Therefore x=4 is an extraneous solution.

You solve the other equations the same way.

anonymous · Answer

I am able to solve 1,2, and 5, but can't get 3 and 4. I looked them up on a calculator and was able to get an answer, but I don't know how to solve them. I couldn't factor them.