Solving for x (quadratic)

[Jules18]

How do you explicitly solve for x in terms of y using this quadratic formula?

y = 2x² + x

I need to solve for x in order to find the inverse fxn. I know how to factor, but after that I'm stuck.
Any suggestions?

~Jules~


* And what about polynomials of a higher order, like x³ + 2x² + x, or something similar?

 

[Andrew Mason]

How do you explicitly solve for x in terms of y using this quadratic formula?

y = 2x² + x

There is no inverse function for y = 2x^2 + x. For some values of y there are more than one value of x (eg. y = 0 or y = 1). So you cannot express x in terms of y.

AM

 

[HallsofIvy]

How do you explicitly solve for x in terms of y using this quadratic formula?

y = 2x² + x

I need to solve for x in order to find the inverse fxn. I know how to factor, but after that I'm stuck.
Any suggestions?

~Jules~


* And what about polynomials of a higher order, like x³ + 2x² + x, or something similar?

Are you not aware of the quadratic formula? If $ax^2+ bx+ c= 0$ then $x= (-b\pm\sqrt{b^2- 4ac})/(2a)$.

In particular, if $y= 2x^2+ x$ then $2x^2+ x- y= 0$ so
$$\frac{-1\pm\sqrt{1+8y}}{4}$$

It is that $\pm$ that prevents this from being a "true" inverse, as Andrew Mason said. We could divide y into two functions, restricting the domain:

If $f_1(x)= 2x^2+ x$ for $-\infty< x\le -1/4$ then
$$f_1^{-1}(x)= \frac{-1-\sqrt{1+8x}{4}$$

If [tex]f_2(x)= 2x^2+ x[/itex] for $-1/4\le x< \infty$ then
$$f_2^{-1}(x)= \frac{-1+ \sqrt{1+8x}{4}$$

There is a general formula for cubics and quartics but they are extremely complicated. It can be shown that cannot be a general formula solve polynomial equations of degree higher than four using only algebraic functions.