Question

Show that ... Has No Real Roots

score 1.8k · Answer 1 · 2026-04-14 02:54:07

1.8k

Views

Show that ... Has No Real Roots

Published on 14 Apr 2026 - 2:54

score 60 · Answer 2 · 2026-03-27 03:56:03

60

Views

Function satisfying inequality has no root

Published on 27 Mar 2026 - 3:56

score 122 · Answer 3 · 2026-04-13 16:55:34

122

Views

Trigonometric Roots of a Polynomial

Published on 13 Apr 2026 - 16:55

score 66 · Answer 4 · 2026-04-08 21:32:41

66

Views

How to compute equation with exponents?

Published on 08 Apr 2026 - 21:32

score 1.5k · Answer 5 · 2026-04-14 09:43:56

1.5k

Views

Complex Conjugate roots with non real coefficients

Published on 14 Apr 2026 - 9:43

score 126 · Answer 6 · 2026-03-29 15:53:09

126

Views

non-complex cubic roots formula?

Published on 29 Mar 2026 - 15:53

score 184 · Answer 7 · 2026-04-19 12:50:29

184

Views

Are there more quadratics with real roots or more with complex roots? Or the same?

Published on 19 Apr 2026 - 12:50

score 73 · Answer 8 · 2026-04-15 01:38:09

73

Views

Prove that $f(x+z)$ has $4$ roots $\pm \alpha$ and $\pm \beta$

Published on 15 Apr 2026 - 1:38

score 120 · Answer 9 · 2026-04-15 11:56:20

120

Views

Question regarding roots of a cubic polynomial

Published on 15 Apr 2026 - 11:56

score 37 · Answer 10 · 2026-04-14 12:45:20

37

Views

Singularity of $f(z)=f(z+a)=f(z+b),\ a,b\in\mathbb C$ does not exist

Published on 14 Apr 2026 - 12:45

score 212 · Answer 11 · 2026-03-26 12:52:50

212

Views

Logistic regression for football results - Estimating coefficient through maximum likelihood

Published on 26 Mar 2026 - 12:52

score 57 · Answer 12 · 2026-04-10 21:39:25

57

Views

Existence of solution of equation involving normal distribution

Published on 10 Apr 2026 - 21:39

score 5.8k · Answer 13 · 2026-03-27 02:50:47

5.8k

Views

Why can a quartic polynomial never have three real and one complex root?

Published on 27 Mar 2026 - 2:50

score 229 · Answer 14 · 2026-04-17 06:46:33

229

Views

Zeros of $f_n(z)=1+\frac{1}{z}+\frac{1}{2!z^2}+...+\frac{1}{n!z^n}$ are in $B_{\varepsilon}(0)$

Published on 17 Apr 2026 - 6:46

score 114 · Answer 15 · 2026-03-29 18:18:39

114

Views

Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$ ,Find out $a, b, c ∈ R$ and its roots knowing that all roots are real.

Published on 29 Mar 2026 - 18:18

List Question

Show that ... Has No Real Roots

Function satisfying inequality has no root

Trigonometric Roots of a Polynomial

How to compute equation with exponents?

Complex Conjugate roots with non real coefficients

non-complex cubic roots formula?

Are there more quadratics with real roots or more with complex roots? Or the same?

Prove that $f(x+z)$ has $4$ roots $\pm \alpha$ and $\pm \beta$

Question regarding roots of a cubic polynomial

Singularity of $f(z)=f(z+a)=f(z+b),\ a,b\in\mathbb C$ does not exist

Logistic regression for football results - Estimating coefficient through maximum likelihood

Existence of solution of equation involving normal distribution

Why can a quartic polynomial never have three real and one complex root?

Zeros of $f_n(z)=1+\frac{1}{z}+\frac{1}{2!z^2}+...+\frac{1}{n!z^n}$ are in $B_{\varepsilon}(0)$

Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$ ,Find out $a, b, c ∈ R$ and its roots knowing that all roots are real.

Trending Questions

Popular # Hahtags

Popular Questions