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Comment by svantana | Hacker News Reader

I feel like this article is a bit backwards - solving quadratic equations is only easy if you have access to the square root function, which by definition is a solver of quadratic expressions. Without it, one needs to resort to iterative root-finding, which works for polynomials of any order.

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That’s not right. For example, if you have access to any sort of radical you still can’t solve the quintic.

"Without it, one needs to resort to iterative root-finding, which works for polynomials of any order."

I believe that's what they mean for quintics. It's been a while for me apologies if I'm miss remembering here.

But if you have access to an extended set of operations (ultraradicals), in particular an operation that solves a parameterized quintic, you can solve all quintics.

https://en.m.wikipedia.org/wiki/Bring_radical

I believe that doesn't extend beyond degree 6 polynomials though. Isn't that Hilbert's 13th problem?

Just use Newton's method!

I can't tell if this is a joke. Newton's method will give you an approximate answer.

You can write the answer as a recurrence in symbolic form.

Or better yet, gpt

Computing (or computing with) a square root is much easier than solving a quadratic equation!

- A square root’s result is always nonnegative (or in the complex case, has nonnegative real part) and unique, whereas the quadratic equation has two solutions.

- You can manipulate the root function symbolically much better than the roots of an equation. sqrt(2)^2 is 2, but figuring out that `solution(x^2-2=0)^2` reduces to the same number is much less obvious, and even more so for more complicated roots.

>A square root’s result is always nonnegative (or in the complex case, has nonnegative real part)

I don’t know what definition you’re using, but if we take the square root of Y to be a number X such that X*X = Y, then the above statement isn’t true.

The principal square root (ie, the one with positive part) is commonly referred to as the square root, especially since the radical symbol is explicitly defined to produce the principal root.

It just seems odd, because the reason why the quadratic equation has/can have two solutions, is due directly to the square root having two solutions.

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