How do you solve `sqrt( 7x^2-1) + 1=3x`?

The first thing you need to do in order to solve this equation is isolate the radical on one side of the equation.

To do that, add `-1` to both sides of the equation

`sqrt(7x""^2 - 1) + cancel(1) - cancel(1) = 3x - 1`

`sqrt(7x""^2 - 1) = 3x-1`

Next, square both sides of the equation to get rid of the radical

`(sqrt(7x""^2 - 1))^2 = (3x-1)^2`

`7x""^2 - 1 = 9x""^2 - 6x + 1`

Move all the terms on one side of the equation

`2x""^2 -6x +2 = 0`

You can use the quadratic formula to determine the two solutions

`x_(1,2) = (6 +- sqrt(36 - 4 * 2 * 2))/4`

`x_(1,2) = (6 +- sqrt(20))/4 = (3 +- sqrt(5))/2 => { (x_1 = (3 + sqrt(5))/2), (x_2 = (3 - sqrt(5))/2) :}`

You need to check these two solutions to see if both are valid, i.e. if you don't have an extraneous solution.

For `x_1` you have

`sqrt(7 * ((3 + sqrt(5))/2)^2-1) = 3 * ((3 + sqrt(5))/2) -1`

This is equivalent to

`sqrt(94 + 42sqrt(5))/cancel(2) = (7 + 3sqrt(5))/cancel(2)`

Square both sides of the equation to get

`(sqrt(94 + 42sqrt(5)))^2 = (7 + 3sqrt(5))^2`

`94 + 42sqrt(5) = 49 + 42sqrt(5) + 45` `->` `x_1` is `color(green)("valid")`

For `x_2` you have

`sqrt(7 * ((3 - sqrt(5))/2)^2-1) = 3 * ((3 - sqrt(5))/2) -1`

This is equivalent to

`94 - 42sqrt(5)) = 49 - 42sqrt(5) + 45` `->` `x_2` is `color(green)("valid")`