Thread: radicals question

I really need help with homework that is due tomorrow, and the problem is:

1. solve (√3x-5) - (√x+7) = 2

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.

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
Say, isolate the 1st radical,
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.

Therefore, x = 18. -------answer.

Thanks guys, your help is greatly appreciated.