Question

Question #8cff4

Answer 1

Isolate one square root, square both sides, the isolate the remaining square root, square again and solve the resulting equation. Check for extraneous solutions.

Explanation:

The condition `x ≥ 1/2` merely makes sure that we are not trying to take the square root of a negative in `sqrt(2x−1)`

We will not use it in the steps below:
`sqrt(x+3)+sqrt(2x−1)=5`

Isolate one of the square roots,. Either will work, I'll isolate the simpler one:

`sqrt(x+3)=5−sqrt(2x−1)`

Now we will square both sides. Carefully!
`(sqrt(x+3))^2 = (5−sqrt(2x−1))^2`

`x+3 = 25−10sqrt(2x−1)+2x-1`

`10sqrt(2x-1) = x+21`

(We could divide by 10 to really isolate the square root, but I'll leave it alone because it is multiplied, not added or subtracted.)

`(10sqrt(2x−1))^2 = (x+21)^2`

`100(2x−1)=x^2+42x+441`

`200x−100=x^2+42x+441`

`x^2−158x+541=0`

Now solve that. Don't spend a lot of time trying to factor, we know we can use the quadratic formula. (it won't factor nicely.)

`x=(−b±sqrt(b^2−4ac))/(2a)`

`=(−(-158) +- sqrt((-158)^2−4(1)(541)))/(2(1))`

`= (158 +- sqrt(22800))/2`
`=(158+- 20sqrt57)/2`
`=79+-10sqrt57`

Now we have to check our answers for extraneous solutions. You'll probably want a calculator to do that, so I'll leave it to you.

`=79+10sqrt57` is extraneous

In brief: `sqrt57` is a little bigger than `7.5`, because `7.5^2 = 6*8+0.25 = 56.25`. So this "solution is near `79+75 = 154` It doesn't take long to be convinced that we won't get `5` if we substitute this on the left of the original equation.

`=79-10sqrt57` solves the equation