Issolvi (x-3)^2(10-17x)^2=0

$$( x - 3 ) ^ { 2 } ( 10 - 17 x ) ^ { 2 } = 0$$

Solvi għal x

$$x=\frac{10}{17}\approx 0.588235294$$

$$x=3$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a-b\right)^{2}=a^{2}-2ab+b^{2} biex tespandi \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(10-17x\right)^{2}=0$$

Uża teorema binomjali \left(a-b\right)^{2}=a^{2}-2ab+b^{2} biex tespandi \left(10-17x\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(100-340x+289x^{2}\right)=0$$

Uża l-propjetà distributtiva biex timmultiplika x^{2}-6x+9 b'100-340x+289x^{2} u kkombina termini simili.

$$4741x^{2}-2074x^{3}+289x^{4}-3660x+900=0$$

Irranġa mill-ġdid l-ekwazzjoni biex tqiegħda fil-forma standard. Qiegħed it-termini f'ordni mill-ogħla qawwa għall-aktar baxxa.

$$289x^{4}-2074x^{3}+4741x^{2}-3660x+900=0$$

Skont it-Teorema tar-Radikali Razzjonali, ir-radikali razzjonali kollha tal-polynomial huma fil-forma \frac{p}{q}, fejn p taqsam il-terminu kostanti 900 u q taqsam il-koeffiċjent prinċipali 289. Elenka l-kandidati kollha \frac{p}{q}.

$$±\frac{900}{289},±\frac{900}{17},±900,±\frac{450}{289},±\frac{450}{17},±450,±\frac{300}{289},±\frac{300}{17},±300,±\frac{225}{289},±\frac{225}{17},±225,±\frac{180}{289},±\frac{180}{17},±180,±\frac{150}{289},±\frac{150}{17},±150,±\frac{100}{289},±\frac{100}{17},±100,±\frac{90}{289},±\frac{90}{17},±90,±\frac{75}{289},±\frac{75}{17},±75,±\frac{60}{289},±\frac{60}{17},±60,±\frac{50}{289},±\frac{50}{17},±50,±\frac{45}{289},±\frac{45}{17},±45,±\frac{36}{289},±\frac{36}{17},±36,±\frac{30}{289},±\frac{30}{17},±30,±\frac{25}{289},±\frac{25}{17},±25,±\frac{20}{289},±\frac{20}{17},±20,±\frac{18}{289},±\frac{18}{17},±18,±\frac{15}{289},±\frac{15}{17},±15,±\frac{12}{289},±\frac{12}{17},±12,±\frac{10}{289},±\frac{10}{17},±10,±\frac{9}{289},±\frac{9}{17},±9,±\frac{6}{289},±\frac{6}{17},±6,±\frac{5}{289},±\frac{5}{17},±5,±\frac{4}{289},±\frac{4}{17},±4,±\frac{3}{289},±\frac{3}{17},±3,±\frac{2}{289},±\frac{2}{17},±2,±\frac{1}{289},±\frac{1}{17},±1$$

Sib radiċi waħda bħal din billi tipprova l-valuri integri kollha, billi tibda mill-iżgħar skont il-valur assolut. Jekk ma tinstab l-ebda radiċi tan-numru integru, ipprova l-frazzjonijiet.

$$x=3$$

Bit-teorema tal-Fattur, x-k hija fattur tal-polynomial għal kull għerq k. Iddividi 289x^{4}-2074x^{3}+4741x^{2}-3660x+900 b'x-3 biex tikseb289x^{3}-1207x^{2}+1120x-300. Solvi l-ekwazzjoni fejn ir-riżultat huwa ugwali għal 0.

$$289x^{3}-1207x^{2}+1120x-300=0$$

Skont it-Teorema tar-Radikali Razzjonali, ir-radikali razzjonali kollha tal-polynomial huma fil-forma \frac{p}{q}, fejn p taqsam il-terminu kostanti -300 u q taqsam il-koeffiċjent prinċipali 289. Elenka l-kandidati kollha \frac{p}{q}.

$$±\frac{300}{289},±\frac{300}{17},±300,±\frac{150}{289},±\frac{150}{17},±150,±\frac{100}{289},±\frac{100}{17},±100,±\frac{75}{289},±\frac{75}{17},±75,±\frac{60}{289},±\frac{60}{17},±60,±\frac{50}{289},±\frac{50}{17},±50,±\frac{30}{289},±\frac{30}{17},±30,±\frac{25}{289},±\frac{25}{17},±25,±\frac{20}{289},±\frac{20}{17},±20,±\frac{15}{289},±\frac{15}{17},±15,±\frac{12}{289},±\frac{12}{17},±12,±\frac{10}{289},±\frac{10}{17},±10,±\frac{6}{289},±\frac{6}{17},±6,±\frac{5}{289},±\frac{5}{17},±5,±\frac{4}{289},±\frac{4}{17},±4,±\frac{3}{289},±\frac{3}{17},±3,±\frac{2}{289},±\frac{2}{17},±2,±\frac{1}{289},±\frac{1}{17},±1$$

Sib radiċi waħda bħal din billi tipprova l-valuri integri kollha, billi tibda mill-iżgħar skont il-valur assolut. Jekk ma tinstab l-ebda radiċi tan-numru integru, ipprova l-frazzjonijiet.

$$x=3$$

Bit-teorema tal-Fattur, x-k hija fattur tal-polynomial għal kull għerq k. Iddividi 289x^{3}-1207x^{2}+1120x-300 b'x-3 biex tikseb289x^{2}-340x+100. Solvi l-ekwazzjoni fejn ir-riżultat huwa ugwali għal 0.

$$289x^{2}-340x+100=0$$

L-ekwazzjonijiet kollha tal-formola ax^{2}+bx+c=0 jistgħu jiġu solvuti billi tuża l-formula kwadratika: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Sostitut 289 għal a, -340 għal b, u 100 għal c fil-formula kwadratika.

$$x=\frac{-\left(-340\right)±\sqrt{\left(-340\right)^{2}-4\times 289\times 100}}{2\times 289}$$

Agħmel il-kalkoli.

$$x=\frac{340±0}{578}$$

Is-soluzzjonijiet huma l-istess.

$$x=\frac{10}{17}$$

Elenka s-soluzzjonijiet kollha misjuba.

$$x=3 x=\frac{10}{17}$$