Question 588089: Problem: Find the inverse of  .
I've fought the good fight, working on this on and off for two weeks.
Status:
For x^2 + x^1/2 = y, the equivalent is x = x^4 - 2x^2y + y^4. I cannot figure out what algebraic manipulative trick(s) to use to separate x, and render it as ONE X term in terms of y. [x = f(y)] Do you have a bag of such applicable tricks?
Since I can't manage that, here's another approach:
x^1/2 = y - x^2 and -(x^1/2) = X^2 - y.
While x^1/2 is not = -(x^1/2), (x^1/2)^2 = x and [-(x^1/2)]^2 = x.
Therefore: x = (y - x^2)^2 and x = (x^2-y)^2.
Now take the square root of each of the above:
x^1/2 = y - x^2 and x^1/2 = x^2 - y.
(Of course, the sq rts can also have a "-" sign.)
Therefore: y - x^2 = x^2 - y.
Is my logic correct so far?
(HOWEVER, in the expression above, let x = 1, and y>1.
If y = 4, 4 - 1 NOT = 1 - 4.)
In any event, if my logic holds up, it seems that I can write:
-2x^2 = -2y AND 2x^2 = 2y.
Each of these is a quadratic. So, it seems that I should be able to use
ax^2 + bx + c = (d)y
inputing known values of x and y [x = 1, y = 2; x = 4, y = 18], and come up with a specific quadratic equation of some form of:
x = -b/2a +/- [(sqrt b^2 + y - 4ac)/2a]
I'll leave it at that. Thanks so much. Cheers!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Problem: Find the inverse of 
:
You seem to have exhausted the usual methods, so I tried this approach
I don't think there is an inverse, anyway
The table for the original equation
x | y
-------
1 | 2
4 |18
9 |84
:
Swapping x & y
x | y
-------
2 | 1
18 | 4
84 | 9
After some tedious math, solving for a, b, c in the y = ax^2+bx+c, I came up with an even more tedious equation
y = -.005452x^2 + .4952x + .5759
A graph kind of shows an inverse, but negative value probably make it invalid
plus the quadratic equation will curve back towards 0
:
After I finished this I thought it was kind of silly, and was not going to bother sending it, but you can just toss in the usual place of rejected stuff.
C
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