Question

Sqrt(2*x+sqrt(6*x^2-1))=x+1

Answer 1

$\sqrt{2x + \sqrt{6 {x}^{2} - 1} } = x + 1$

$( { \sqrt{2x + \sqrt{6 {x}^{2} - 1 } } )}^{2} = ({x + 1})^{2}$

$2x + \sqrt{6 {x}^{2} - 1 } = {x}^{2} + 2x + 1$

$\sqrt{6 {x}^{2} - 1 } = {x}^{2} + 1$

$( { \sqrt{6 {x}^{2} - 1 }) }^{2} = {x}^{4} + 2 {x}^{2} + 1$

$6 {x}^{2} - 1 = {x}^{4} + {x}^{2} + 1$

${x}^{4} - 5 {x}^{2} + 2 = 0$

Answer 2

To solve the equation Sqrt(2x + sqrt(6x^2 - 1)) = x + 1, we'll follow these steps:

1. Simplify the equation by eliminating the square root on the left side.
2. Isolate the square root term.
3. Square both sides of the equation to eliminate the remaining square root(s).
4. Solve the resulting quadratic equation for x.
5. Check the solutions for extraneous roots.

Let's proceed with these steps:

1. Simplify the equation:

Start with Sqrt(2x + sqrt(6x^2 - 1)) = x + 1 Square both sides: (Sqrt(2x + sqrt(6x^2 - 1)))^2 = (x + 1)^2 2x + sqrt(6x^2 - 1) = x^2 + 2*x + 1

2. Isolate the square root term:

Subtract x^2 and 2x from both sides: sqrt(6*x^2 - 1) = x^2 + 1

3. Square both sides again:

(√(6x^2 - 1))^2 = (x^2 + 1)^2 6x^2 - 1 = x^4 + 2*x^2 + 1

4. Rearrange to form a quadratic equation:

x^4 + 2x^2 - 6x^2 + 1 + 1 = 0 x^4 - 4*x^2 + 2 = 0

5. Solve the quadratic equation:

Unfortunately, this is a quartic equation, which can be quite challenging to solve directly. There is no simple algebraic solution for quartic equations like there is for quadratic or cubic equations.

One way to proceed is to use numerical methods to approximate the roots. You can use numerical solvers or graphing tools to find approximate solutions. Additionally, it's possible that this equation might not have real solutions, and it could have complex roots.

Please note that the complexity of this equation highlights the importance of considering the context in which it arises. If this is part of a larger problem or scenario, there might be alternative methods to approach the overall problem.

If you have a specific value of x or more context related to the problem, I can try to provide further assistance.