SOLUTION: Here is my problem. [SQRT(x + 7)] - 2[SQRT(x)] =-2 OR √(x + 7) - 2√(x) = -2 ------- Now, the steps I took to solve this problem are to first square both

Algebra ->  Radicals -> SOLUTION: Here is my problem. [SQRT(x + 7)] - 2[SQRT(x)] =-2 OR √(x + 7) - 2√(x) = -2 ------- Now, the steps I took to solve this problem are to first square both      Log On


   

Question 720317: Here is my problem.
[SQRT(x + 7)] - 2[SQRT(x)] =-2
OR
√(x + 7) - 2√(x) = -2
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
So then I thought to subtract (x+7) and (4x) to both sides
- [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7 (I think I'm supposed to switch the sign, because I've subtracted it and moved it to the opposite side, right?)
I think I'm correct up to this point, but now I have to square both sides again.
I think this left hand side could be re-written as:
-2[√(x + 7) * - 2√(x)]
Is this right? I'm subtracting it from itself, a negative, which could simply multiplied by -2. Anyway, now I need to square this again, so I assume the -2 becomes a 4 and I FOIL them separately?
FOILING the left side will get:
[√(x + 7) * - 2√(x)] * [√(x + 7) * - 2√(x)]
Which, when FOILed, looks like
(x+7) - [√(x+7) * -2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I want to know if figuring this one out could help me solve the other ones.
I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot of other things.
One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original equation and it works, but I'm just confused about how to actually get 9 out of this..
Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all the steps of solving a problem like this so I can do it easily in the future.
Found 3 solutions by lwsshak3, MathTherapy, greenestamps:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
√(x + 7) - 2√(x) = -2
√(x+7)=2√x-2
square both sides
x+7=4x-8√x+4
-3x+3=-8√x
-3(x-1)=-8√x
square both sides again
9(x-1)^2=64x
9(x^2-2x+1)=64x
9x^2-18x+9-64x=0
9x^2-82x+9=0
solve for x using quadratic formula:
x≈0.1111..(reject, extraneous root)
x=9

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Here is my problem.

[SQRT(x + 7)] - 2[SQRT(x)] =-2   OR   √(x + 7) - 2√(x) = -2 
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
So then I thought to subtract (x+7) and (4x) to both sides
 - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7 (I think I'm supposed to switch the sign, because I've
subtracted it and moved it to the opposite side, right?)
I think I'm correct up to this point, but now I have to square both sides again.
I think this left hand side could be re-written as:
 -2[√(x + 7) * - 2√(x)]
Is this right? I'm subtracting it from itself, a negative, which could simply multiplied by -2. Anyway, now I need to
square this again, so I assume the -2 becomes a 4 and I FOIL them separately?
FOILING the left side will get:
 [√(x + 7) * - 2√(x)] * [√(x + 7) * - 2√(x)]
Which, when FOILed, looks like
(x+7) - [√(x+7) * -2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I
want to know if figuring this one out could help me solve the other ones.

I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot
of other things.
One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original
equation and it works, but I'm just confused about how to actually get 9 out of this..

Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all
the steps of solving a problem like this so I can do it easily in the future.
*******************************
Here is my problem.
[SQRT(x + 7)] - 2[SQRT(x)] =-2      OR       √(x + 7) - 2√(x) = -2 
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x <=== This is where you went WRONG! When FOILed, this's
actually: , which results in:
                 x+%2B+7+-+2sqrt%28x%29sqrt%28x+%2B+7%29+-+2sqrt%28x%29sqrt%28x+%2B+7%29+%2B+4x+=+4. See?          
                                     x+%2B+7+-+4sqrt%28x%29sqrt%28x+%2B+7%29+%2B+4x+=+4                                                                                        
                                                    -+4sqrt%28x%29sqrt%28x+%2B+7%29+=+4+-+%28x+%2B+7%29+-+4x  So then I thought to subtract (x+7) and (4x) to both sides
                                                    -+4sqrt%28x%29sqrt%28x+%2B+7%29+=+4+-+x+-+7+-+4x%29  - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7
                                                                                                     (I think I'm supposed to switch the sign, because I've 
                                                                                                     subtracted it and moved it to the opposite side, right?)
                                                  -+4sqrt%28x%5E2+%2B+7x%29+=+-+3+-+5x <=== This is exactly how you need to proceed!
                                           %28-+4sqrt%28x%5E2+%2B+7x%29%29%5E2+=+%28-+3+-+5x%29%5E2 ---- Squaring both sides
                                                 16%28x%5E2+%2B+7x%29+=+9+%2B+30x+%2B+25x%5E2
                                                  16x%5E2+%2B+112x+=+9+%2B+30x+%2B+25x%5E2
                       16x%5E2+%2B+112x+-+9+-+30x+-+25x%5E2+=+0
                                             -+9x%5E2+%2B+82x+-+9+=+0        OR       9x%5E2+-+82x+%2B+9+=+0
                                                                                            9x%5E2+-+81x+-+x+%2B+9+=+0
                                                                                       9x(x - 9) - 1(x - 9) = 0
                                                                                               (x - 9)(9x - 1) = 0
                                                                                                x - 9 = 0               OR      9x - 1 = 0
                                                                                                      x = 0 + 9        OR           9x = 1
                                                                                                      x = 9               OR              x = 1%2F9 (IGNORE)
1%2F9 proves to be EXTRANEOUS, so sole solution is: x = 9
To some though, it's easier to solve, if one of the left-side RADICALS is MOVED to the right, 1st. But then, this's subjective.


It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I want
to know if figuring this one out could help me solve the other ones. Hopefully, the above will clear up SOME/ALL confusion!

I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot of
other things.  Hopefully, the above will clear up SOME/ALL confusion!

One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original equation
and it works, but I'm just confused about how to actually get 9 out of this.  Yes, the solution is indeed 9, as seen above! 

Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all the steps
of solving a problem like this so I can do it easily in the future.  Hopefully, this author has assisted you in understanding this 
problem, so you can understand and obtain SOLUTIONS to similar problems, more easily, more efficiently, and without confusion!

Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Your original equation is sqrt%28x%2B7%29-2sqrt%28x%29=-2

You CAN start the problem by squaring both sides of the equation; but when you do that you end up with an equation involving an "uglier" square root. Reaching the final answer is almost always easier if you start by changing the equation so that there is only one radical on each side of the equation.

And in this particular example, with the "-2" on the right side, I would also, to avoid possible future confusion, move that term to the left hand side, making the equation

sqrt%28x%2B7%29%2B2=2sqrt%28x%29

Now square both sides.

%28x%2B7%29%2B4sqrt%28x%2B7%29%2B4=4x

Now isolate the remaining radical and square both sides again.

4sqrt%28x%2B7%29=3x-11
16%28x%2B7%29=9x%5E2-66x%2B121
16x%2B112=9x%5E2-66x%2B121
9x%5E2-82x%2B9=0
%289x-1%29%28x-9%29=0

x=1%2F9 or x=9

We squared both sides of the equation in solving the problem, so some of the roots we ended up with might not satisfy the original equation, so we need t check. x = 1/9 does NOT satisfy the original equation; x = 9 does. So the unique solution to the given equation is x = 9.

ANSWER: x = 9


A final note....

I work with high school math students who often take timed competitive tests where the speed of getting the answer is important. While the intent here is almost certainly to solve the problem by formal mathematical methods, the problem can be solved informally VERY quickly using logical reasoning.

The original equation shows that a combination via addition or subtraction of the radicals sqrt(x+7) and sqrt(x) yields in an integer result. That means (x+7) and (x) must both be integer perfect squares. A little knowledge of perfect square integers immediately tells us that the only two perfect square integers that differ by 7 are 16 and 9.

So we quickly have our solution:
ANSWER: x = 9