SOLUTION: (4x+5)^.5-(x+4)^.5=(x-1)^.5

Question 631114: (4x+5)^.5-(x+4)^.5=(x-1)^.5
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Since you categorized this problem as a problem of radicals, you must already know the .5 (or 1/2) as an exponent means square root. So I'm going to rewrite the equation with square roots:

Here's a procedure to solve these kinds of equations:

Isolate a square root.
Square both sides of the equation. (Note: Squaring the isolated square root should be easy. But be careful with the other side of the equation.)
If there are still some square roots, repeat steps 1, 2 and 3.
At this point all the square roots should be gone. Determine the type of equation you now have (linear, quadratic, other) and use appropriate techniques to solve it.
Check you solution(s). This is not optional! Squaring both sides of the equation has occurred at least once to get this far. This is not a wrong thing to do. But squaring both sides of an equation can introduce what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation! Extraneous solutions can occur even if no mistakes have been made. Even expert mathematicians have to check these answers. If you have made no mistakes but you find a solution that does not fit the original equation then reject/discard it.

Let's see this in action on your equation:
1. Isolate a square root.
The square root on the right is already isolated. But I am going to add to each side anyway. This will isolate and it will eliminate a subtraction. (Mistakes are easier to make with subtractions than additions.)

2. Square both sides.

Squaring the left side is easy. Squaring the right side is not as easy at it might look. Exponents do not distribute. To square the right side, we must either use FOIL on or use the pattern with the "a" being "" and the "b" being "". I prefer to use the patterns:

Simplifying...

3. We still have a square root so we repeat.

1. Isolate a square root.
There is only one square root left. Subtracting 2x and 3 from each side we get:

The 2 in front of the square root is not a problem. Squaring the right side will still be easy. But if it bothers you, you can divide both sides by 2 to get rid of it.

2. Square both sides.

The right side squares easily and we can use the pattern again to square the left side:

Simplifying:

3. The square roots are gone! SO we can proceed to step 4.

4. Solve the equation.
This may look like a quadratic equation. But if we subtract from each side, squared terms will both disappear:

This is a linear equation. Subtracting 8x from each side:

Adding 16:

Dividing both sides by 4:

5. Check your solution(s).
Use the original equation to check:

Checking x = 5:

Simplifying...

Check!
So x = 5 is the only solution to your equation.

P.S. If x = 5 had not checked out and if we had not made any mistakes, then x = 5 would have been an extraneous solution. We would have had to reject it and. since it was our only possible solution, there would have been no solutions to the equation!
RELATED QUESTIONS
(x� -4x -5)/(x� +2x +1) � (x -5)/(x... (answered by nerdybill)

-(x-1)+4x =... (answered by checkley75)

4x(x-5)-5x(x-4)=-1 (answered by clarkp)

1- (x-5)/5 = (x+6)/4 - (4x-1)/20 (answered by stanbon)

x(x+5)/4x+4=9/5 (answered by Fombitz)

5(x+1)-x=4x+5 (answered by Fombitz)

5/2x+1/4x=9/4+x (answered by Fombitz)

5/2x + 1/4x = 7/4 +... (answered by rfer)