Imonyula umsuka wenombolo eyinkimbinkimbi

Kulolu shicilelo, sizobheka ukuthi ungathatha kanjani impande yenombolo eyinkimbinkimbi, nokuthi lokhu kungasiza kanjani ekuxazululeni izibalo ze-quadratic okubandlulula kwazo kungaphansi kukaziro.

Imonyula umsuka wenombolo eyinkimbinkimbi

I-Square Root

Njengoba sazi, akunakwenzeka ukuthatha umsuka wenombolo yangempela eyinegethivu. Kodwa uma kukhulunywa ngezinombolo eziyinkimbinkimbi, lesi senzo singenziwa. Ake sikuthole.

Ake sithi sinenombolo z = -9. Ngokuba √-9 kunezimpande ezimbili:

z₁ =√-9 = -3i

z₁ =√-9 = 3i

Ake sihlole imiphumela etholiwe ngokuxazulula i-equation z² =-9, ngingakukhohlwa lokho i² =-1:

(-3i)² = (-3)² ⋅ i² = 9 ⋅ (-1) = -9

(3i)² = 3² ⋅ i² = 9 ⋅ (-1) = -9

Ngakho, sikufakazile lokho -3i и 3i ziyizimpande √-9.

Umsuka wenombolo enegethivu ngokuvamile ubhalwa kanje:

√-1 = ±i

√-4 = ±2i

√-9 = ±3i

√-16 = ±4i njll

Impande emandleni n

Ake sithi sinikezwe izibalo zefomu z = ⁿ√w… Ine n izimpande (z₀, of₁, of₂,…, z_n-1), engabalwa kusetshenziswa ifomula engezansi:

|w| iyimojuli yenombolo eyinkimbinkimbi w;

φ - impikiswano yakhe

k ipharamitha ethatha amanani: k = {0, 1, 2,…, n-1}.

Izibalo zequadratic ezinezimpande eziyinkimbinkimbi

Ukukhipha impande yenombolo engalungile kushintsha umqondo ojwayelekile we-uXNUMXbuXNUMXb. Uma ubandlululo (D) ingaphansi kukaziro, ngakho-ke ngeke kube khona izimpande zangempela, kodwa zingamelwa njengezinombolo eziyinkimbinkimbi.

Isibonelo

Masixazulule isibalo x² - 8x + 20 = 0.

Isixazululo

a = 1, b = -8, c = 20

D = b² – 4ac = 64 – 80 = -16

D <0, kodwa sisengakwazi ukuthatha umsuka wokubandlulula okubi:

√D =√-16 = ±4i

Manje singakwazi ukubala izimpande:

x_1,2 = (-b ± √D)/2a = (8 ± 4i)/2 = 4 ±2i.

Ngakho-ke, i-equation x² - 8x + 20 = 0 inezimpande ezimbili eziyinkimbinkimbi ze-conjugate:

x₁ = 4 + 2i

x₂ = 4 – 2i