CalcOpenly
मराठी

बहुपदीची मुळे शोधा

All real and complex roots of a polynomial up to degree 6, such as a cubic or quartic equation, with repeated roots found exactly and 30-digit accuracy.

अद्यतनित तपासलेली उदाहरणे: ७

1, -6, 11, -6 means x³ − 6x² + 11x − 6. Include zeros for missing powers.
वापरून पाहा
Roots
1, 2, 3
Roots: 1, 2, 3
Real roots (counted with multiplicity)
3
Largest real root
3
Smallest real root
1
Degree
3

This degree-3 polynomial has 3 real roots (counting multiplicity).

p(x) = x³ − 6x² + 11x − 6

-2-1012123xp(x)x = 1x = 2x = 3

Roots in the complex plane (dashed: unit circle)

ReIm123
Roots ओळी: ३
#Real partImaginary partMultiplicity|p(root)|
11010
22010
33010
गणना कशी केली जाते S

  1. Polynomial

    p(x)=x3−6x2+11x−6p(x) = x^{3} - 6x^{2} + 11x - 6

  2. Durand–Kerner iteration

    zk←zk−p(zk)∏j≠k(zk−zj)z_k \leftarrow z_k - \frac{p(z_k)}{\prod_{j \ne k}(z_k - z_j)}

    Started from powers of 0.4 + 0.9i scaled by the Cauchy bound; 12 sweeps until every correction was below 10⁻³⁰ of the root, then Newton polishing.

  3. Exact rational roots

    x=1,x=2,x=3x = 1,\quad x = 2,\quad x = 3

    Each was confirmed by substituting the fraction into p(x) and getting exactly 0.

  4. Check with Vieta's formulas

    ∑zk=6=−an−1an=6,∏zk=6=(−1)na0an=6\sum z_k = 6 = -\tfrac{a_{n-1}}{a_n} = 6,\qquad \prod z_k = 6 = (-1)^{n}\tfrac{a_0}{a_n} = 6

बहुपदीची मुळे शोधा बद्दल

A polynomial of degree n has exactly n roots among the complex numbers, counted with multiplicity; this is the fundamental theorem of algebra. The calculator first splits off repeated factors exactly with Yun's square-free algorithm, then finds all remaining roots at once with the Durand–Kerner (Weierstrass) iteration, which refines n guesses together until each correction is below 10⁻³⁰ of the root. Rational roots are confirmed by exact substitution.

Cubic and quartic equations from engineering, physics and algebra courses are typical inputs. The default, 1, −6, 11, −6, is x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3), so the roots are 1, 2 and 3; they sum to 6 and multiply to 6, as Vieta's formulas require.

Enter coefficients from the highest power down, with zeros for missing powers: x³ − 2 is 1, 0, 0, −2. Degrees 1 to 6 are accepted.

सोडवलेली उदाहरणे

x³ − 6x² + 11x − 6

Coefficients, highest power first
1, -6, 11, -6
Roots
1, 2, 3
Real roots (counted with multiplicity)
3
Largest real root
3
Smallest real root
1

पडताळणीचा स्रोत: (x − 1)(x − 2)(x − 3) expanded by hand

x³ − 2 (one real, two complex)

Coefficients, highest power first
1, 0, 0, -2
Roots
1.25992105, −0.6299605249 ± 1.091123636i
Real roots (counted with multiplicity)
1
Largest real root
1.2599210499

पडताळणीचा स्रोत: Python decimal: 2^(1/3) and 2^(1/3)·(−1/2 ± i√3/2)

x⁴ − 1

Coefficients, highest power first
1, 0, 0, 0, -1
Roots
−1, 1, ±i
Real roots (counted with multiplicity)
2

पडताळणीचा स्रोत: Fourth roots of unity: ±1, ±i

Repeated root (x − 1)³(x + 2)

Coefficients, highest power first
1, -1, -3, 5, -2
Roots
−2, 1 (×3)
Real roots (counted with multiplicity)
4
Smallest real root
-2

पडताळणीचा स्रोत: (x − 1)³(x + 2) = x⁴ − x³ − 3x² + 5x − 2 (expanded with Python fractions)

प्रश्न

How many roots does a polynomial have?

Exactly as many as its degree, counted with multiplicity, once complex roots are included; this is the fundamental theorem of algebra. x⁴ − 1 has four roots: −1, 1, i and −i. The number of real roots can be smaller: x³ − 2 has one real root, ∛2 ≈ 1.259921, and two complex ones. A root of multiplicity 3, such as x = 1 in (x − 1)³(x + 2), counts three times.

How do you solve a cubic equation?

Look for a rational root first. By the rational root theorem, any rational root p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient. For x³ − 6x² + 11x − 6, trying divisors of 6 finds x = 1, and dividing by (x − 1) leaves x² − 5x + 6 = (x − 2)(x − 3). Without a rational root, Cardano's formula or a numerical method is needed.

Is there a formula for the roots of a quintic?

No general formula using radicals exists for degree 5 or higher. The Abel–Ruffini theorem, proved by Abel in 1824, shows this, and Galois theory explains which equations can be solved that way. x⁵ − x − 1 is a standard example whose roots cannot be written with radicals. Numerical methods still find them: its only real root is about 1.167304.

What are Vieta's formulas?

They link the roots to the coefficients. For aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀, the roots add up to −aₙ₋₁/aₙ and multiply to (−1)ⁿa₀/aₙ. For x³ − 6x² + 11x − 6 the roots 1, 2 and 3 sum to 6 and multiply to 6, matching −(−6)/1 and (−1)³ × (−6)/1. The calculator uses both as a check on the roots it finds.

Why do complex roots come in conjugate pairs?

When every coefficient is real, conjugating the equation p(z) = 0 gives p(z̄) = 0, so the conjugate of a root is also a root. x³ − 2 therefore has the pair −0.629961 ± 1.091124i alongside its real root. It follows that a polynomial of odd degree with real coefficients always has at least one real root.

“बहुपदीची मुळे शोधा” किती अचूक आहे?

अचूकता तुमच्या इनपुटवर आणि पद्धतीच्या गृहीतकांवर अवलंबून असते. दशांश गणना 50 सार्थ अंक वापरते, पण अंदाज, संख्यात्मक पद्धती आणि मूळ डेटा कमी अचूक असू शकतात; दाखवलेल्या मूल्यांचे पूर्णांकन केल्याने या मर्यादा दूर होत नाहीत. स्वतंत्र स्रोतांतील सोडवलेल्या उदाहरणांशी पडताळणी: ७. उदाहरणार्थ, “x³ − 6x² + 11x − 6” ची पडताळणी (x − 1)(x − 2)(x − 3) expanded by hand याच्याशी केली आहे.

या पद्धतीचा स्रोत कोणता?

Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration); D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976.

या कॅल्क्युलेटरबद्दल

zk←zk−p(zk)∏j≠k(zk−zj)z_k \leftarrow z_k - \frac{p(z_k)}{\prod_{j \ne k} (z_k - z_j)}

स्रोत

  1. Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration)
  2. D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976

संदर्भांशी पडताळलेले

या कॅल्क्युलेटरमध्ये स्वतंत्र स्रोतांमधील उत्तरांसह ७ सोडवलेली उदाहरणे आहेत. ती चाचणी संचात चालतात आणि तुम्ही ती येथेही चालवू शकता.

संबंधित कॅल्क्युलेटर