solve x+x^(-1/2)-6=0 over real numbers

Question

solving quadratic over real numbers

Answer 1

x+x^-½-6=0,

x+1/√x-6=0,

x√x+1-6√x=0.

Let y=√x, then x=y², so y³-6y+1=0.

This cubic has 3 real roots: -2.52892, 0.16745, 2.36147, from which x can be calculated by squaring each of these values.

But how do we find the roots of a cubic? There are various methods. Here's one.

A cubic equation always has at least one real root. Using Newton's iterative method:

y_n+1=y_n-f(y_n)/f'(y_n), where f(y)=y³-6y+1 and f'(y)=3y²-6. Let y₀=0 then x₁=-1/(-6)=⅙=0.1666...

y₁=107/639=0.167449..., y₂=0.1674491911, y₃=0.167449191109 (stable value). x=0.02804 approx.

So one solution to the cubic is y=0.167449191109. Let this be denoted by A. Using synthetic division we can deduce a quadratic:

A | 1 0     -6          1

     1 A A²    | A³-6A

     1 A A²-6 |     0       =y²+Ay+A²-6. This quadratic has two solutions:

y=½(-A±√(24-3A²))=2.361468766 or -2.528917957.

So x=5.5765 or 6.3954 approx.

CHECK:

Substituting each of these three values into the original equation we get:

x=5.5765: x+x^-½-6=0✔️; x=6.3954: x+x^-½-6=0.7909❌; x=0.02804: x+x^-½-6=0✔️

So there are two solutions: x=5.5765 and x=0.02804. The rejected answer is caused by assuming x^-½ can be negative as well as positive.