Question

How do you solve for k in `d + 3 = sqrt(2k – 5)`?

Answer 1

`k=(d^2+6d+14)/2`

Explanation:

`d+3=sqrt(2k-5)`

we can exchange the order off the factors:

`sqrt(2k-5)=d+3`

Now square both sides of the equation. sqrt(2k-5) is a positive number so we must add the constriction d+3>=0. Otherwise we would add a false solution:

`(sqrt(2k-5))^2=(d+3)^2 and d+3>=0`

`2k-5=d^2+6d+9 and d>= -3`

`2k=d^2+6d+14 and d>= -3`

`k=(d^2+6d+14)/2 and d>= -3`