(2x-40)^(1/2) + (x-80)^(1/2) = 30?

Question

I can't get this at all. I don't know what I can and can't do here.

Answer 1

First step: study when this equation exists

2x >=40, so x >= 20

And

And x>=80

So, x >= 80

When the square roots exists, they are positive or 0

Now lets solve this

(2x-40)^(1/2) = 30 - (x-80)^(1/2)

We dont know if the right side is positive or not. If it is, since the left side is non negative, you can square. But, if its not, you will be introducing roots. So, lets square and then lets check if we introduced roots or not.

2x - 40 = 900 - 60 (x-80)^(1/2) + x - 80

60 (x-80)^(1/2) = - x + 780

Lets square again.

3600 (x-80) = x^2 - 2*780x + 780^2

Find x and verify if these roots are roots from the original equation

Ana

Answer 2

(2x - 40)^(1/2) + (x - 80)^(1/2) = 30
Squaring both sides you have
2x - 40 + 2((2x - 40)(x - 80))^(1/2) + x - 80 = 900
2((2x - 40)(x - 80))^(1/2) + 3x - 120 = 900
2((2x - 40)(x - 80))^(1/2) = 780 - 3x
((2x - 40)(x - 80))^(1/2) = (780 - 3x)/2
Squaring again you have
(2x - 40)(x - 80) = (780 - 3x)^2/4
4(2x^2 - 200x + 3200) = 608400 - 2820x + 9x^2
8x^2 - 800x + 12800 = 608400 - 2820x + 9x^2
x^2 - 2,020x + 595,600 = 0
x = (2,020 ± √(4,080,400 - 2,382,400))/2
x = (2,020 ± √1,698,000)/2
x = 1,010 ± 20√4245