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I am reading in the Visual Group Therapy book that the above equation is unsolvable algebraically. What are other ways to solve this if not through algebra?

Evan Kim
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    "Solvable" (i.e., solvable by radicals) is (presumably) a term of art here: It means that the zeros of the polynomial can be obtained from integers applying the four arithmetic operations and the operation of taking $n$th roots (as in the quadratic formula). Short of that, there are still numeric approximation methods, convenient closed forms in terms of special functions, etc. – anomaly Mar 10 '19 at 04:22

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The roots of $x^5+10x^4-2=0$ can be computed approximately by various numerical methods.

They are three real roots : $x\simeq -9.9998 \quad;\quad -0.68063\quad;\quad 0.658168$

and two complex roots $x\simeq 0.0111307\pm 0.66809\,i $.

$x^5+10x^4-2=0$ is a quintic equation.

In general (except particular cases) the quintic equation cannot be solved in terms of a finite number of elementary functions (Abel and Galois theory). But it is solvable in terms of special functions, namely the Jacobi theta functions.

http://mathworld.wolfram.com/QuinticEquation.html

JJacquelin
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There are many ways to estimate the roots of a polynomial (which I assume you are talking about). A typical method is to solve them numerically, say using Newton's method. There are other numerical schemes, but I'd argue that this is the most elementary one - usually taught in a first calculus course or definitely in a first numerical analysis course.

More generally, I believe Newton's method falls under something known as fixed point iteration. On this page, you'll find other techniques that can be used to estimate roots of an equation.

Sorey
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