Question

How do you solve `sqrt(2x+1/6)=x+5/6`?

Answer 1

I have taken you to a point where you can finish it off.

Explanation:

If you have a square root on one side of the equals and a 'non' square root on the other, you can 'get rid' of the root by squaring both sides. Giving:

`2x+1/6=(x+5/6)^2`

`2x+1/6=x^2+5/3x+25/36`

Collecting terms on one side only of the equals to give us a quadratic equation.

`x^2+5/3x-2x+25/36-1/6=0`

`x^2-1/3x +19/36=0`

Compare to `y=ax^2+bx+c` where `x=(-b+-sqrt(b^2-4ac))/(2a)`

`=> x=(1/3+-sqrt((1/9)-4(1)(19/36)))/(2(1))`

I will let you finish this off.