Mathematician Finds Easier Way to Solve Quadratic Equations

[Albert Chan]

(05-01-2024 04:57 PM)Gerson W. Barbosa Wrote:  Given the second degree equation

\(x^2+px+q=0\)

... Then

\(x_{1}=\sqrt{q}\times e^{\cosh^{-1}{(-\frac{p}{2\sqrt{q}}})}\)

The second root can be computed as

\(x_{1} x_{2}= q\)

\(x_{2}=\frac{q}{x_{1}}\)

Nice! And, we can make formula more compact, 2 roots with ±

(08-01-2021 03:08 PM)Albert Chan Wrote:  We can also solve for z² - 2*m*z + n² = 0

XCas> m,n := (-18+5i)/-2., sqrt(45.-15i)
XCas> proot([1, -2m, n*n])                                    → [3.0, 15.0-5.0*i]
XCas> n*tan(asin(n/m)/2), n/tan(asin(n/m)/2)        → [3.0, 15.0-5.0*i]
XCas> n/exp(acosh(m/n)), n*exp(acosh(m/n))        → [3.0, 15.0-5.0*i]

Again, combine both roots, z = n*exp(±acosh(m/n))

tan/asin quadratic formula based from tan half-angle formula:

sin(x) = 2t/(1+t^2), where t = tan(x/2)
t^2 - 2*(1/sin(x))*t + 1 = 0

exp/acosh quadratic formula based from cosh definition.

cosh(x) = (y + 1/y)/2, where y = e^x
y^2 - 2*cosh(x)*y + 1 = 0

Above patterns matched with scaled z² - 2*m*z + n² = 0

(z/n)^2 - 2*(m/n)*(z/n) + 1 = 0