Question

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

Answer 1

This equation has 2 solutions: `x_1=4-2sqrt(2)` and `x_2=4+2sqrt(2)`

Explanation:

In the equation there is a square root, so first you have to calculate the domain.

In this case the experssion under root sign is `2x+1` so you have the domain: `2x+1>=0` or `x>=-1/2`

Now we can calculate the solutions.
We start with raising both sides to the second power

`2x+1=(x-3)^2`

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

`-x^2+8x-8=0`

Now we can solve the quadratic equation:

`Delta = 8^2-4*(-1)*(-8)`

`Delta=64-32=32`

`sqrt(Delta)=sqrt(32)=4sqrt(2)`

`x_1=(-8-4sqrt(2))/(2*(-1))`

`x_1=4+2sqrt(2)`

`x_2=(-8+4sqrt(2))/(2*(-1))`

`x_2=4-2sqrt(2)`

Now we have to check if any of calculated solutions are in the domain.

`x_1>4` so it surely is greater than `-1/2`

`x_2=4-2sqrt(2) ~~4-2*1,41~~4-2,82~~1,18>=-1/2`

We checked that both solutions are in the domain, so we can write the answer:

This equation has 2 solutions: `x_1=4-2sqrt(2)` and `x_2=4+2sqrt(2)`