# math

how would you solve for y in this problem:
x=(y^2+3y)^(1/3)
I got: y= x^3/y - 3 ?

and for this problem:
x= 1/[(y^2+3y-5)^3]
I got: y = y/x + 5/y - 3?

However, I do not know how to solve for y explicitly. I don't think quadratic formula would work in these problems. Please help me out!

OK let's look at the first one.
Your equation is correct but does no allow one to calculate y directly.

y^2 + 3y = x^3
You can use the quadratic formula to solve for y explicitly, treating -x^3 as the constant "c" term in the general expression
ay^2 + by + c = 0
This results in
y = [-3 +/- sqrt (9 + 4x^3]/2

Now for the second one
(1/x)^(1/3) = y^2 + 3y -5
The term on the left if the cube root of 1/x.
Again, treat -5 - (1/x)^(1/3) as the "c" term in the general quadratic equation,. and solve for y
y = -6 +/- sq

Some suggestions kristie:
Graph the function y^2 + 3y first. You'll see that it's not 1-1, i.e. vertical lines intersect the funcion in more than 1 point.
When you take the 3rd root of it, it's still not 1-1.
this means we have to restrict the domain.
In your sol'n we don't divide by y. Cube both sides, complete the square for y, take roots and identify the branches for the inverse function.
So we have X^3 = y^2+3y, then x^3 + 9/4 = y^2 +3y +9/4 = (y + 3/2)^2
Take roots and substract 3/2. There will be two branches here. Check my work.

what did you mean by 1-1?

I think I mentioned this below.
By 1-1 (read one-one) we mean a function such that any horizontal line intersects the function in at most one point. If you graph y = x^2, you can see that any horizontal line above the x-axis intersects the function twice. This is because (-x)^2 = (+x)^2, thus it is not 1-1. However, y=x^3 is 1-1. Graph it and see. In order for a function to have an inverse that is not 'split apart', it must be 1-1.

Note:I may have said "vertical" line when I originally used the term. It should be any "horizontal" line.

No, when we solve for x we don't want y in any of our terms. What we're trying to do is 'undo' all the operations on y so it stands alone.
Here we start by reciprocating both sides.
We have x^(-1) = y^2+3y-5)^3
Now take 3rd roots, again solve to get 3rd root(x^(-1))=(y-a)^2 (I'll let you figure out what a is.) Take roots, and isolate y. Does that make sense?

how did you get x^(-1)? I thought the reciprocal of 1 is 1.

I reciprocrated the x: reciprocal(x) = 1/x = x^(-1)

BTW, I hope this doesn't confuse things, but you can use the interchange and solve for the other variable method for all your inverse problems. Thus for this one,
x= 1/[(y^2+3y-5)^3] interchange gives,
y= 1/[(x^2+3x-5)^3] and solve for x, so
1/y = [(x^2+3x-5)^3], then
(1/y)^(1/3) = x^2+3x-5, then
(1/y)^(1/3)+5 = x^2+3x,
(1/y)^(1/3)+5 +9/4 = x^2+3x+9/4,
(1/y)^(1/3)+5 +9/4 = (x+3/2)^2
+/- root((1/y)^(1/3)+5 +9/4) = (x+3/2)
+/- root((1/y)^(1/3)+5 +9/4)-3/2 = x
Then simplify by doing the addition.
You can solve this one fairly straight-forward too, just thought I'd mention this.
__________________
Also, I think in the other problem I said solve for x when it should have been solve for y; check that.

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