Question

Find the value(s) of x in the following radical equations:

1. √X-3 = 5
4. 13x - 1 - √x+ 5 = 0
2. √x+8=3
5. x = √x - 4+4
3. 3√2x + 3 + 5 = 0
6. x + 4 = x + 10

Answer 1

The values of x in the given radical equations are as follows:

1. x = 2916
2. x = 16
3. No real solutions
4. x = 7
5. No real solutions
6. No real solutions

Step-by-step explanation:

In the first equation, √x - 3 = 5, we can solve for x by isolating the square root term and then squaring both sides to eliminate the square root. This yields x = 54 + 3^2, which simplifies to x = 2916.

Moving on to the second equation, 13x - 1 - √x + 5 = 0, we can isolate the square root term and then square both sides to eliminate it. After simplifying, we find that x = 16.

For the third equation, √x + 8 = 35, isolating the square root term and squaring both sides gives us x = 35² - 8, which simplifies to x = 1225 - 8, resulting in x = 1217. However, upon further examination, this solution is not valid as it leads to a negative value under the square root, indicating no real solutions.

In the fourth equation, x = √x - 4 + 4, we can simplify to get x - √x + 4 = 0. By substituting y = √x, we obtain a quadratic equation y^2 - y - 4 = 0. Solving for y gives us two complex roots, which means there are no real solutions for x.

For the fifth equation, 3√2x + 3 + 5 = 0 simplifies to √2x + 8 = 0. Isolating the square root term and squaring both sides results in a contradiction as it leads to a negative value under the square root, indicating no real solutions.

Lastly, in the sixth equation, x + 4 = x + 10 simplifies to a contradiction where both sides are equal but also not equal at the same time, indicating no real solutions.