Question 720317: Here is my problem.
[SQRT(x + 7)] - 2[SQRT(x)] =-2
OR
√(x + 7) - 2√(x) = -2
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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.
Answer by lwsshak3(11628)   (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
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