y^2 -2y = x^3 + 2x^2 + 2x + 3 solve this for y in terms of x (book says its a simple quadratic in y?)?

Question

I need to understand why you get the answer y = +/- root(x^3 + 2x^2 + 2x + 4) My differential textbook does not explain this and I have forgotten my first year calculus, thanks.

Thanks for all of your help, that is exactly what I needed.

Answer 1

Completing the square:
y^2 - 2y = y^2 - 2y + 1 - 1 = (y - 1)^2 - 1

y^2 -2y = x^3 + 2x^2 + 2x + 3

(y - 1)^2 - 1 = x^3 + 2x^2 + 2x + 3 // + 1

(y - 1)^2 = x^3 + 2x^2 + 2x + 4 // Take square roots

y - 1 = +- sqrt(x^3 + 2x^2 + 2x + 4) // + 1

y = +- sqrt(x^3 + 2x^2 + 2x + 4) + 1

Answer 2

Hai TREVOR B,

First things First.

This has nothing do with Calculus. This is simple algebra!

A quadratic means any expression having 2 as the largest power of the variable, and is expressed in the form ax^2 + bx + c, where, x is the variable and a, b & c are constants.

In the featured problem, x is y, a is 1, b is (-2) & c is 0(zero).

Next, the answer you get will be y = +/- root(x^3 + 2x^2 + 2x + 4) + 1

Finally, how you get the answer has been correctly shown by Amit Y.

I have tried to simplify it further for better understanding:

We have, y^2 -2y = x^3 + 2x^2 + 2x + 3

Adding 1 to both LHS & RHS, we get, y^2 -2y+1 = x^3 + 2x^2 + 2x + 3+1

i. e. y^2 -2y+1 = x^3 + 2x^2 + 2x + 4 .....(A)

We know that, y^2 -2y+1 = (y - 1)^2 .....(B)

Using (B) in (A), we get, (y - 1)^2 = x^3 + 2x^2 + 2x + 4

Taking square roots, we get, (y - 1) = +/- root(x^3 + 2x^2 + 2x + 4)
Hence, we get, y = [+/- root(x^3 + 2x^2 + 2x + 4)]+1

Hope this is helpful.

Enjoy your mathematics!

Answer 3

You can treat the "x" terms as the constant in the Quadratic Formula:

ay² + by + c = 0

y² - 2y - (x³ + 2x² + 2x + 3) = 0

Use the quadratic formula to get to the next step:

y = [2 +/- root(4 - 4(1)(x³ + 2x² + 2x + 3))] / 2

Factor out a 4 from the root, and it becomes 2, which cancels out with the denominator:

y = 1 +/- root(1 - (x³ + 2x² + 2x + 3))

Answer 4

y^2 - 2y = x^3 + 2x^2 + 2x + 3

Complete the square.

y^2 - 2y + 1 = x^3 + 2x^2 + 2x + 4
(y - 1)^2 = x^3 + 2x^2 + 2x + 4
(y - 1)^2 = (x + 2)*(x^2 + 2)