Could someone help me solve this?
(2x-1)^1/2 - (x)^1/2 = 2
Here’s one way: move the second term with the square root from the left side to the right side. Then square both sides. Then, move all the terms without radicals back to the left side, clean up and then square both sides again! You will get a quadratic equation that actually factors nicely…but wait: squaring both sides of an equation often introduces “extraneous” solutions…so check both of your answers to see which one works.
I hope that makes sense. If you need more detail, let me know.
I tried your suggestion but because I end up with a sum that includes a square root of a number, I can’t seem to get rid of the square root and therefore have trouble factoring.
A friend suggested substituting y for square root of x, which allows me to factor and then squaring the solution for y. Do you think that makes sense?
Assuming x is a non-negative real:
sqrt(2x-1) = 2 + sqrt(x)
Square both sides:
2x-1 = 4 + 4 sqrt(x) + x
<==> 4 sqrt(x) = x-5
Square both sides again:
16x = x^2 - 10x + 25
<==> x^2 - 26x + 25 = 0
<==> (x-25)(x-1) = 0
x = 25 or x = 1
However x = 1 is an extraneous solution as it was introduced when squaring again, but x = 25 works, which is the only real solution.
Got it. I think you did what pckeller suggested and it makes sense. Better than using y as my friend suggested.
Thanks…
One more question. How do I solve:
sqrt x + sqrt (x+5) = 5 (sqrt x-3)
@CHD2013 x = 4 is a solution.
A good rule of thumb for evaluating all real solutions is to square both sides and isolate the square root term, so that when you square again, you are left with a polynomial. A pitfall is that squaring both sides often adds extraneous solutions, because x^2 = y^2 does not imply x = y. As @pckeller said, you’ll have to check whether your answers actually work.
Another method is to find one (or more) solutions such as x = 4 and then somehow prove that it is the only solution. For example, you can prove that x = 25 is the only real solution by showing that the function f(x) = sqrt(2x-1) - sqrt(x) (where x >= 1/2) has a positive derivative everywhere.
OK, but finding that solution…well, you can start by playing around with numbers, knowing that the person who wrote the problem probably picked the solution first! So it is probably a nice perfect square that leads to other nice squares…
But as for: sqrt x + sqrt (x+5) = 5 (sqrt x-3)
You just have to do more of same: square both sides, get the radical term on one side, clean up, square both sides again, clean up again…regroup and hope that it factors.
These are more fun to write than to solve. For example, say I want to write one of these where the answer is 7. I notice that 7+2 is a perfect square and so is 7-3. So if I want a mildly annoying question:
sqrt(x+2) - sqrt(x-3) = 1
Not annoying enough? 7+ 9 is also a perfect square:
sqrt(x+9) + sqrt(x-3) = 2sqrt(x+2)
Still not annoying enough? Well, (4)(7) - 3 is also a perfect square…and so is (2)(7)+2.
So sqrt(4x-3) = sqrt(2x+2) + sqrt(x-6)
OK, I’ll stop now.
@pckeller heh. This somehow reminds me of a pretty neat algebra problem, from the Stanford Math Tournament.
Q: Find all real x such that cube root of (20x + cube root of (20x+13)) = 13.
(i.e. $\sqrt[3]{20x + \sqrt[3]{20x+13}} = 13$ if you know LaTeX)