find all numbers x € R satisfying:?

Question

i get how for |2x+1|=|2=x|, x=1/3 or -3, but then i dont get the other examples. how do you find all numbers where x is real for the following??

|x²+5x-40|=|x²+3x-24|

| |x-1| - |x+1| | = x²

[[2x - 1]] = 3

[[x + 2]] = 2[[x]] - 3

[[x]] = [[2x]]

yes, the double brackets are for greatest integer.

Answer 1

The first two problems are easy enough, and have been solved for you, so here is a method which will do the rest - provided I've interpreted what you have said correctly.


Problem 3: ||2x - 2|| = 3

Let x = N + α, where N is an integer with N ≤ x, and 0 ≤ α < 1

Then for ||2x - 1|| = 3, we have || 2(N + α) - 1 || = 3
so ||2N -1 + 2α|| = 3

Note that the 2N - 1 is an integer, so all that matters is the 2α bit.
If 0 ≤ α < 0.5, 0 ≤ 2α < 1, so ||2N -1 + 2α|| = 2N -1 ∴ 2N - 1 = 3 ⇒ N = 2
⇒ x = 2 + α, so 2 ≤ x < 2.5

If 0.5 ≤ α < 1, then 1 ≤ 2α < 2, so ||2N -1 + 2α|| = 2N ∴ 2N = 3, but this is impossible as N is an integer, so there are no additional solutions.

∴ 2 ≤ x < 2.5


Problem 4: ||x + 2|| = 2||x|| - 3

Again, let x = N + α, where N is an integer with N ≤ x, and 0 ≤ α < 1

then ||N + 2 + α|| = 2||N + α|| - 3, so, as 0 ≤ α < 1, we must have:

N + 2 = 2N - 3 ⇒ 5 = N, so x = 5 + α,
∴ solution is 5 ≤ x < 6


Problem 5: ||x|| = ||2x||

Again, let x = N + α, where N is an integer with N ≤ x, and 0 ≤ α < 1,

then ||N + α|| = ||2N + 2α|| . . .once again, the important value of α occurs when 2α = 1

So, when 0 ≤ α < 0.5, we will have N = 2N, so N = 0
∴ the solutions are x = 0 + α ⇒ 0 ≤ x < 0.5

when 0.5 ≤ α < 1, we must have ||2N + 2α|| = 2N + 1,
so the solutions occur where N = 2N + 1, so N = -1
∴ the solutions are x = -1 + α ⇒ -0.5 ≤ x < 0

The overall solution is therefore -0.5 ≤ x < 0.5

I hope you find the method useful.

Answer 2

I forget what the double brackets means (if it's "greatest integer <=, in my day I think we used single brackets for that).

Anyhow, for the first one there are two possibilites: The quadratics equal each other, or they equal each other's negative.

The first possibility gives you 2x = 16. The second one gives you 2x^2 + 8x - 64 = 0. Surely you can solve both of those. :)

The second one is trickier. To keep the number of possibilities from going out of control, note this:

||a|-|b|| = |a-b| if a and b have the same sign (or one is 0), and equals |a+b| if they have the opposite sign (or one is 0)

Well, in the first case we're looking at 2 = x^2. That has two roots. And yes, x-1 and x=1 have the same sign in each case, so they are both solutions.

In the second case, we have |2x| = x^2. That gives two subcases. One is where 2x=x^2. That has roots 2 and 0. But 2 does NOT satisfy the condition of x-1 and x+1 having opposite signs. So the only real solution we get here is x=0. In the other subcase, we get roots x=-2 and x=0 again. And x=-2 is also a bogus solution.

So the three solutions to the second problem are 0, sqrt(x), and -sqrt(2).