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I'm trying to inverse this equation as a function of $x$

$$z = x + \frac{x^2}{2}$$

but couldn't wrap up my head around it. If anyone can provide a step by step solution to this it will be really helpful.

Thanks in advance!

Pedro
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clinraus
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  • Try completing the square - i.e. writing the right hand side as $(x + a)^2$ for some appropriate $a$. Also, inversion works on 1-1 functions, so you'd have to restrict your domain appropriately. – Utku Alhan Mar 18 '14 at 14:04

3 Answers3

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To be strict, the "inverse" of a function can only exist if the function is one-one. Hence, strictly speaking, an inverse of the given function does not exist, because there are multiple values of $x$ which maps to the same value of $z$.

But just for educational purposes : rearrange to get

$$x^2 + 2x - 2z = 0$$

Now, this is just your regular quadratic equation. Recall how to solve for $x$ using the quadratic formula $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. Apply this to find $x$ in terms of $z$ (notice how I am not calling this the "inverse"). You could also "complete the square", whichever you are more comfortable with.

Yiyuan Lee
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Rewrite

$$x^2+2x=2z$$ then $$(x+1)^2=2z+1.$$ You should be able to conclude.

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In general, you canot invert the equation you want. For example, if $x=1$, then $z = \frac32$. The problem is that if $x=-3$, you also have $z=-3+\frac92=\frac32$. So, what will the inverse of $\frac32$ be?

5xum
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