• Sep 8th 2005, 05:46 PM
u317d
I really need help with homework that is due tomorrow, and the problem is:

1. solve (√3x-5) - (√x+7) = 2
• Sep 8th 2005, 06:37 PM
MathGuru
Quote:

Originally Posted by u317d
I really need help with homework that is due tomorrow, and the problem is:

1. solve (√3x-5) - (√x+7) = 2

Do you mean √(3x-5) - √(x+7) = 2 ?

If so I would graph this equation on the left hand side and then graph y=2 and find where it intersects.
• Sep 9th 2005, 11:03 AM
ticbol
solve (√3x-5) - (√x+7) = 2

If that is
sqrt(3x -5) -sqrt(x+7) = 2,
then,....

In cases like this, you have to isolate one radical to one side of the equation, then rationalize that isolated radical--and of course the other side of the equation too, at the same time--so that the radical sign will be eliminated. Then keep on isolating radicals and rationalizing them until you don't have anymore radicals left.

Here, the radicals are all square roots, so you just square a square root to rationalize it.

Too much talk, eh?
Let us do it.

sqrt(3x -5) -sqrt(x+7) = 2
sqrt(3x -5) = 2 +sqrt(x+7)
Square both sides,
(3x -5) = 2^2 +2*2*sqrt(x+7) +(x+7)
3x -5 = 4 +4sqrt(x+7) +x +7
We have one more radical left, so we isolate that again,
3x -5 -4 -x -7 = 4sqrt(x+7)
2x -16 = 4sqrt(x+7)
Divide both sides by 2,
x -8 = 2sqrt(x+7)
Square both sides,
x^2 -16x +64 = 4(x+7)
x^2 -16x +64 = 4x +28
x^2 -16x +64 -4x -28 = 0
x^2 -20x +36 = 0
(x -18)(x-2) = 0

x-18 = 0
x = 18 ---------***

x-2 = 0
x = 2 ------------***

Check if those satisfy the original "sqrt(3x -5) -sqrt(x+7) = 2"

When x=18,
sqrt(3*18 -5) -sqrt(18+7) =? 2
sqrt(49) -sqrt(25) =? 2
7 -5 =? 2
2 =? 2
Yes, so, OK.

When x=2
sqrt(3*2 -5) -sqrt(2+7) =? 2
sqrt(1) -sqrt(9) =? 2
1 -3 =? 2
-2 =? 2
No, so x=2 is not a solution.