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15
Math.TechQA.Club
2014-12-06 15:50:29
405
Views
Conditioning of the calculation of roots for cubic polynomial
Published on
06 Dec 2014 - 15:50
#numerical-methods
#roots
594
Views
Show all roots of $\sum_{k=0}^n 2^{k(n-k)} x^k$ are real (December 6, 2014 Putnam problem)
Published on
06 Dec 2014 - 21:43
#real-analysis
#polynomials
#summation
#roots
46
Views
Solving for $x$ using $\ln$ or any possible way.
Published on
06 Dec 2014 - 22:46
#algebra-precalculus
#logarithms
#roots
120
Views
Applying Newton-Raphson method to $a\cdot b^{-2}=c\cdot d^4+e\cdot f(d)$
Published on
30 Mar 2026 - 7:20
#algebra-precalculus
#functions
#numerical-methods
#roots
#approximation
464
Views
Sum of square of absolute values of roots of a polynomial
Published on
08 Dec 2014 - 2:00
#linear-algebra
#polynomials
#roots
78
Views
Solving $3t^2-\frac{12}{3}t+\frac{4}{3}=0$
Published on
01 Apr 2026 - 21:08
#algebra-precalculus
#polynomials
#roots
#factoring
220
Views
$f'(a)=0$ implies $x=a$ is not a simple zero of $f$
Published on
08 Dec 2014 - 18:00
#derivatives
#polynomials
#proof-writing
#roots
51
Views
Find a solution for an equation
Published on
08 Dec 2014 - 18:18
#roots
139
Views
Roots of a polynomial equation where coefficients follow a geometric progression
Published on
09 Dec 2014 - 1:56
#linear-algebra
#complex-analysis
#polynomials
#roots
90
Views
Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$
Published on
09 Dec 2014 - 12:14
#algebra-precalculus
#roots
10.8k
Views
how many distinct real zeros a function has
Published on
09 Dec 2014 - 18:45
#polynomials
#roots
71
Views
Find the equation which has key root $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$
Published on
26 Mar 2026 - 19:17
#roots
#symmetric-polynomials
2.6k
Views
$x^4 + x^2 + 1 = 0$ has no solution in $\mathbb{R}$.
Published on
11 Dec 2014 - 21:00
#roots
61
Views
A problem related to complex polynomial
Published on
12 Dec 2014 - 5:12
#algebra-precalculus
#complex-analysis
#polynomials
#roots
1.2k
Views
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Published on
30 Mar 2026 - 8:00
#algebra-precalculus
#trigonometry
#polynomials
#roots
#radicals
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