Question

Solve the radical equation √2-√36 - 2x = 6. Check for extraneous solutions.

The solution is |
The extraneous solution is
Enter the real solution first followed by the extarneous solution separated
by a comma.
For example, if real solution is 10 and the extraneous solution is 2.
Enter as 10,2.

Answer 1

To solve the radical equation √2-√36 - 2x = 6, isolate the radical term by simplifying the equation and then solving for x. Check for extraneous solutions by substituting the value of x back into the original equation and evaluating if it is true. In this case, there are no extraneous solutions.

Step-by-step explanation:

To solve the radical equation √2-√36 - 2x = 6, we start by simplifying the equation. The square root of 36 is 6, so the equation becomes √2 - 6 - 2x = 6. Next, we isolate the radical term by subtracting 6 from both sides of the equation, which gives us √2 - 2x = 12. To solve for x, we isolate the radical term again by subtracting √2 from both sides, giving us -2x = 12 - √2. Finally, we divide both sides by -2 to solve for x, which gives us x = (12 - √2)/-2. This is the real solution.

Now, we need to check for extraneous solutions. We substitute the value of x back into the original equation and simplify. If the simplified equation is not true, then we have an extraneous solution. In this case, when we substitute x = (12 - √2)/-2 back into the original equation, we get √2 - √36 - 2((12 - √2)/-2) = 6. Simplifying this equation, we get 6 = 6, which is true. Therefore, there are no extraneous solutions.

Answer 2

We have the equation:

√2 - √36 - 2x = 6

Simplifying the radicals we get:

√2 - 6 - 2x = 6

Adding 6 to both sides we get:

√2 - 2x = 12

Subtracting √2 from both sides we get:

-2x = 12 - √2

Dividing by -2 we get:

x = (6 - (1/√2))

Therefore, the real solution is:

x = (6 - (1/√2))

To check for extraneous solutions, we need to substitute this value of x back into the original equation and check if it satisfies the equation.

√2 - √36 - 2(6 - (1/√2)) = 6

Simplifying this we get:

-6 = 6

This is not true, so the solution x = (6 - (1/√2)) is extraneous.

Therefore, there is no real solution to the equation.