Riješi 125(2x-1)^3+2=66

$$125 ( 2 x - 1 ) ^ { 3 } + 2 = 66$$

Izračunaj x (complex solution)

$$x=\frac{9}{10}=0,9$$

$$x=\frac{\sqrt{3}i}{5}+\frac{3}{10}\approx 0,3+0,346410162i$$

$$x=-\frac{\sqrt{3}i}{5}+\frac{3}{10}\approx 0,3-0,346410162i$$

Koraci rješenja

Upotrijebite binomni teorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} da biste proširili \left(2x-1\right)^{3}.

$$125\left(8x^{3}-12x^{2}+6x-1\right)+2=66$$

Koristite svojstvo distributivnosti da biste pomnožili 125 s 8x^{3}-12x^{2}+6x-1.

$$1000x^{3}-1500x^{2}+750x-125+2=66$$

Dodajte -125 broju 2 da biste dobili -123.

$$1000x^{3}-1500x^{2}+750x-123=66$$

Oduzmite 66 od obiju strana.

$$1000x^{3}-1500x^{2}+750x-123-66=0$$

Oduzmite 66 od -123 da biste dobili -189.

$$1000x^{3}-1500x^{2}+750x-189=0$$

Prema teoremu racionalnog korijena, svi racionalni korijeni polinomijalnog oblika su u obliku \frac{p}{q}, gdje p dijeli konstantni termin -189 i q dijeli glavni koeficijent 1000. Navedite sve kandidate \frac{p}{q}.

$$±\frac{189}{1000},±\frac{189}{500},±\frac{189}{250},±\frac{189}{200},±\frac{189}{125},±\frac{189}{100},±\frac{189}{50},±\frac{189}{40},±\frac{189}{25},±\frac{189}{20},±\frac{189}{10},±\frac{189}{8},±\frac{189}{5},±\frac{189}{4},±\frac{189}{2},±189,±\frac{63}{1000},±\frac{63}{500},±\frac{63}{250},±\frac{63}{200},±\frac{63}{125},±\frac{63}{100},±\frac{63}{50},±\frac{63}{40},±\frac{63}{25},±\frac{63}{20},±\frac{63}{10},±\frac{63}{8},±\frac{63}{5},±\frac{63}{4},±\frac{63}{2},±63,±\frac{27}{1000},±\frac{27}{500},±\frac{27}{250},±\frac{27}{200},±\frac{27}{125},±\frac{27}{100},±\frac{27}{50},±\frac{27}{40},±\frac{27}{25},±\frac{27}{20},±\frac{27}{10},±\frac{27}{8},±\frac{27}{5},±\frac{27}{4},±\frac{27}{2},±27,±\frac{21}{1000},±\frac{21}{500},±\frac{21}{250},±\frac{21}{200},±\frac{21}{125},±\frac{21}{100},±\frac{21}{50},±\frac{21}{40},±\frac{21}{25},±\frac{21}{20},±\frac{21}{10},±\frac{21}{8},±\frac{21}{5},±\frac{21}{4},±\frac{21}{2},±21,±\frac{9}{1000},±\frac{9}{500},±\frac{9}{250},±\frac{9}{200},±\frac{9}{125},±\frac{9}{100},±\frac{9}{50},±\frac{9}{40},±\frac{9}{25},±\frac{9}{20},±\frac{9}{10},±\frac{9}{8},±\frac{9}{5},±\frac{9}{4},±\frac{9}{2},±9,±\frac{7}{1000},±\frac{7}{500},±\frac{7}{250},±\frac{7}{200},±\frac{7}{125},±\frac{7}{100},±\frac{7}{50},±\frac{7}{40},±\frac{7}{25},±\frac{7}{20},±\frac{7}{10},±\frac{7}{8},±\frac{7}{5},±\frac{7}{4},±\frac{7}{2},±7,±\frac{3}{1000},±\frac{3}{500},±\frac{3}{250},±\frac{3}{200},±\frac{3}{125},±\frac{3}{100},±\frac{3}{50},±\frac{3}{40},±\frac{3}{25},±\frac{3}{20},±\frac{3}{10},±\frac{3}{8},±\frac{3}{5},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{1000},±\frac{1}{500},±\frac{1}{250},±\frac{1}{200},±\frac{1}{125},±\frac{1}{100},±\frac{1}{50},±\frac{1}{40},±\frac{1}{25},±\frac{1}{20},±\frac{1}{10},±\frac{1}{8},±\frac{1}{5},±\frac{1}{4},±\frac{1}{2},±1$$

Pronađite takav korijen tako da isprobate sve cjelobrojne vrijednosti, počevši od najmanje apsolutne vrijednosti. Ako se ne pronađu cjelobrojni korijeni, pokušajte s razlomcima.

$$x=\frac{9}{10}$$

Faktor teorem, x-k je faktor polinoma za svaki korijenski k. Podijelite 1000x^{3}-1500x^{2}+750x-189 s 10\left(x-\frac{9}{10}\right)=10x-9 da biste dobili 100x^{2}-60x+21. Riješite jednadžbu u kojoj rezultat odgovara 0.

$$100x^{2}-60x+21=0$$

Sve jednadžbe oblika ax^{2}+bx+c=0 mogu se riješiti pomoću kvadratne formule: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. U kvadratnoj formuli zamijenite 100 s a, -60 s b i 21 s c.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 100\times 21}}{2\times 100}$$

Izračunajte.

$$x=\frac{60±\sqrt{-4800}}{200}$$

Riješite jednadžbu 100x^{2}-60x+21=0 kad je ± plus i kad je ± minus.

$$x=-\frac{\sqrt{3}i}{5}+\frac{3}{10} x=\frac{\sqrt{3}i}{5}+\frac{3}{10}$$

Navedi sva pronađena rješenja.

$$x=\frac{9}{10} x=-\frac{\sqrt{3}i}{5}+\frac{3}{10} x=\frac{\sqrt{3}i}{5}+\frac{3}{10}$$

Izračunaj x

$$x=\frac{9}{10}=0,9$$

Koraci rješenja

Upotrijebite binomni teorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} da biste proširili \left(2x-1\right)^{3}.

$$125\left(8x^{3}-12x^{2}+6x-1\right)+2=66$$

Koristite svojstvo distributivnosti da biste pomnožili 125 s 8x^{3}-12x^{2}+6x-1.

$$1000x^{3}-1500x^{2}+750x-125+2=66$$

Dodajte -125 broju 2 da biste dobili -123.

$$1000x^{3}-1500x^{2}+750x-123=66$$

Oduzmite 66 od obiju strana.

$$1000x^{3}-1500x^{2}+750x-123-66=0$$

Oduzmite 66 od -123 da biste dobili -189.

$$1000x^{3}-1500x^{2}+750x-189=0$$

Prema teoremu racionalnog korijena, svi racionalni korijeni polinomijalnog oblika su u obliku \frac{p}{q}, gdje p dijeli konstantni termin -189 i q dijeli glavni koeficijent 1000. Navedite sve kandidate \frac{p}{q}.

$$±\frac{189}{1000},±\frac{189}{500},±\frac{189}{250},±\frac{189}{200},±\frac{189}{125},±\frac{189}{100},±\frac{189}{50},±\frac{189}{40},±\frac{189}{25},±\frac{189}{20},±\frac{189}{10},±\frac{189}{8},±\frac{189}{5},±\frac{189}{4},±\frac{189}{2},±189,±\frac{63}{1000},±\frac{63}{500},±\frac{63}{250},±\frac{63}{200},±\frac{63}{125},±\frac{63}{100},±\frac{63}{50},±\frac{63}{40},±\frac{63}{25},±\frac{63}{20},±\frac{63}{10},±\frac{63}{8},±\frac{63}{5},±\frac{63}{4},±\frac{63}{2},±63,±\frac{27}{1000},±\frac{27}{500},±\frac{27}{250},±\frac{27}{200},±\frac{27}{125},±\frac{27}{100},±\frac{27}{50},±\frac{27}{40},±\frac{27}{25},±\frac{27}{20},±\frac{27}{10},±\frac{27}{8},±\frac{27}{5},±\frac{27}{4},±\frac{27}{2},±27,±\frac{21}{1000},±\frac{21}{500},±\frac{21}{250},±\frac{21}{200},±\frac{21}{125},±\frac{21}{100},±\frac{21}{50},±\frac{21}{40},±\frac{21}{25},±\frac{21}{20},±\frac{21}{10},±\frac{21}{8},±\frac{21}{5},±\frac{21}{4},±\frac{21}{2},±21,±\frac{9}{1000},±\frac{9}{500},±\frac{9}{250},±\frac{9}{200},±\frac{9}{125},±\frac{9}{100},±\frac{9}{50},±\frac{9}{40},±\frac{9}{25},±\frac{9}{20},±\frac{9}{10},±\frac{9}{8},±\frac{9}{5},±\frac{9}{4},±\frac{9}{2},±9,±\frac{7}{1000},±\frac{7}{500},±\frac{7}{250},±\frac{7}{200},±\frac{7}{125},±\frac{7}{100},±\frac{7}{50},±\frac{7}{40},±\frac{7}{25},±\frac{7}{20},±\frac{7}{10},±\frac{7}{8},±\frac{7}{5},±\frac{7}{4},±\frac{7}{2},±7,±\frac{3}{1000},±\frac{3}{500},±\frac{3}{250},±\frac{3}{200},±\frac{3}{125},±\frac{3}{100},±\frac{3}{50},±\frac{3}{40},±\frac{3}{25},±\frac{3}{20},±\frac{3}{10},±\frac{3}{8},±\frac{3}{5},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{1000},±\frac{1}{500},±\frac{1}{250},±\frac{1}{200},±\frac{1}{125},±\frac{1}{100},±\frac{1}{50},±\frac{1}{40},±\frac{1}{25},±\frac{1}{20},±\frac{1}{10},±\frac{1}{8},±\frac{1}{5},±\frac{1}{4},±\frac{1}{2},±1$$

Pronađite takav korijen tako da isprobate sve cjelobrojne vrijednosti, počevši od najmanje apsolutne vrijednosti. Ako se ne pronađu cjelobrojni korijeni, pokušajte s razlomcima.

$$x=\frac{9}{10}$$

Faktor teorem, x-k je faktor polinoma za svaki korijenski k. Podijelite 1000x^{3}-1500x^{2}+750x-189 s 10\left(x-\frac{9}{10}\right)=10x-9 da biste dobili 100x^{2}-60x+21. Riješite jednadžbu u kojoj rezultat odgovara 0.

$$100x^{2}-60x+21=0$$

Sve jednadžbe oblika ax^{2}+bx+c=0 mogu se riješiti pomoću kvadratne formule: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. U kvadratnoj formuli zamijenite 100 s a, -60 s b i 21 s c.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 100\times 21}}{2\times 100}$$

Izračunajte.

$$x=\frac{60±\sqrt{-4800}}{200}$$

Kvadratni korijen negativnog broja nije definiran u polju realnih brojeva, pa nema rješenja.

$$x\in \emptyset$$

Navedi sva pronađena rješenja.

$$x=\frac{9}{10}$$