Resolver 0^2=(25x^2-60x-36)(25x^2+60x+36)

$$0 ^ { 2 } = ( 25 x ^ { 2 } - 60 x - 36 ) ( 25 x ^ { 2 } + 60 x + 36 )$$

Ebatzi: x

$$x = \frac{6 \sqrt{2} + 6}{5} \approx 2.897056275$$

$$x=\frac{6-6\sqrt{2}}{5}\approx -0.497056275$$

$$x = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

Ebazpidea

0 lortzeko, egin 0 ber 2.

$$0=\left(25x^{2}-60x-36\right)\left(25x^{2}+60x+36\right)$$

Erabili banaketa-propietatea 25x^{2}-60x-36 eta 25x^{2}+60x+36 biderkatzeko eta antzeko gaiak bateratzeko.

$$0=625x^{4}-3600x^{2}-4320x-1296$$

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$625x^{4}-3600x^{2}-4320x-1296=0$$

Erro arrazionalaren teoremari jarraikiz, polinomioen erro arrazional guztiek \frac{p}{q} forma dute, non p balioak -1296 balio konstantea zatitzen duen, eta q balioak 625 koefiziente nagusia zatitzen duen. Zerrendatu hautagai guztiak \frac{p}{q}.

$$±\frac{1296}{625},±\frac{1296}{125},±\frac{1296}{25},±\frac{1296}{5},±1296,±\frac{648}{625},±\frac{648}{125},±\frac{648}{25},±\frac{648}{5},±648,±\frac{432}{625},±\frac{432}{125},±\frac{432}{25},±\frac{432}{5},±432,±\frac{324}{625},±\frac{324}{125},±\frac{324}{25},±\frac{324}{5},±324,±\frac{216}{625},±\frac{216}{125},±\frac{216}{25},±\frac{216}{5},±216,±\frac{162}{625},±\frac{162}{125},±\frac{162}{25},±\frac{162}{5},±162,±\frac{144}{625},±\frac{144}{125},±\frac{144}{25},±\frac{144}{5},±144,±\frac{108}{625},±\frac{108}{125},±\frac{108}{25},±\frac{108}{5},±108,±\frac{81}{625},±\frac{81}{125},±\frac{81}{25},±\frac{81}{5},±81,±\frac{72}{625},±\frac{72}{125},±\frac{72}{25},±\frac{72}{5},±72,±\frac{54}{625},±\frac{54}{125},±\frac{54}{25},±\frac{54}{5},±54,±\frac{48}{625},±\frac{48}{125},±\frac{48}{25},±\frac{48}{5},±48,±\frac{36}{625},±\frac{36}{125},±\frac{36}{25},±\frac{36}{5},±36,±\frac{27}{625},±\frac{27}{125},±\frac{27}{25},±\frac{27}{5},±27,±\frac{24}{625},±\frac{24}{125},±\frac{24}{25},±\frac{24}{5},±24,±\frac{18}{625},±\frac{18}{125},±\frac{18}{25},±\frac{18}{5},±18,±\frac{16}{625},±\frac{16}{125},±\frac{16}{25},±\frac{16}{5},±16,±\frac{12}{625},±\frac{12}{125},±\frac{12}{25},±\frac{12}{5},±12,±\frac{9}{625},±\frac{9}{125},±\frac{9}{25},±\frac{9}{5},±9,±\frac{8}{625},±\frac{8}{125},±\frac{8}{25},±\frac{8}{5},±8,±\frac{6}{625},±\frac{6}{125},±\frac{6}{25},±\frac{6}{5},±6,±\frac{4}{625},±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{3}{625},±\frac{3}{125},±\frac{3}{25},±\frac{3}{5},±3,±\frac{2}{625},±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{625},±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

Aurkitu halako erro bat osoko balio guztiak probatuta, balio txikienetik hasita, eta balio absolutuak erabiliz. Ez baduzu aurkitzen osoko errorik, probatu zatikiak.

$$x=-\frac{6}{5}$$

Biderkagaien teoremari jarraikiz, polinomioaren biderkagai bat da x-k, k erro bakoitzeko. 125x^{3}-150x^{2}-540x-216 lortzeko, zatitu 625x^{4}-3600x^{2}-4320x-1296 5\left(x+\frac{6}{5}\right)=5x+6 balioarekin. Ebatzi ekuazioa, hura eta 0 berdinak izan arte.

$$125x^{3}-150x^{2}-540x-216=0$$

Erro arrazionalaren teoremari jarraikiz, polinomioen erro arrazional guztiek \frac{p}{q} forma dute, non p balioak -216 balio konstantea zatitzen duen, eta q balioak 125 koefiziente nagusia zatitzen duen. Zerrendatu hautagai guztiak \frac{p}{q}.

$$±\frac{216}{125},±\frac{216}{25},±\frac{216}{5},±216,±\frac{108}{125},±\frac{108}{25},±\frac{108}{5},±108,±\frac{72}{125},±\frac{72}{25},±\frac{72}{5},±72,±\frac{54}{125},±\frac{54}{25},±\frac{54}{5},±54,±\frac{36}{125},±\frac{36}{25},±\frac{36}{5},±36,±\frac{27}{125},±\frac{27}{25},±\frac{27}{5},±27,±\frac{24}{125},±\frac{24}{25},±\frac{24}{5},±24,±\frac{18}{125},±\frac{18}{25},±\frac{18}{5},±18,±\frac{12}{125},±\frac{12}{25},±\frac{12}{5},±12,±\frac{9}{125},±\frac{9}{25},±\frac{9}{5},±9,±\frac{8}{125},±\frac{8}{25},±\frac{8}{5},±8,±\frac{6}{125},±\frac{6}{25},±\frac{6}{5},±6,±\frac{4}{125},±\frac{4}{25},±\frac{4}{5},±4,±\frac{3}{125},±\frac{3}{25},±\frac{3}{5},±3,±\frac{2}{125},±\frac{2}{25},±\frac{2}{5},±2,±\frac{1}{125},±\frac{1}{25},±\frac{1}{5},±1$$

Aurkitu halako erro bat osoko balio guztiak probatuta, balio txikienetik hasita, eta balio absolutuak erabiliz. Ez baduzu aurkitzen osoko errorik, probatu zatikiak.

$$x=-\frac{6}{5}$$

Biderkagaien teoremari jarraikiz, polinomioaren biderkagai bat da x-k, k erro bakoitzeko. 25x^{2}-60x-36 lortzeko, zatitu 125x^{3}-150x^{2}-540x-216 5\left(x+\frac{6}{5}\right)=5x+6 balioarekin. Ebatzi ekuazioa, hura eta 0 berdinak izan arte.

$$25x^{2}-60x-36=0$$

ax^{2}+bx+c=0 erako ekuazio guztiak formula koadratiko honen bidez ebatz daitezke: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Ordeztu 25 balioa a balioarekin, -60 balioa b balioarekin, eta -36 balioa c balioarekin formula koadratikoan.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 25\left(-36\right)}}{2\times 25}$$

Egin kalkuluak.

$$x=\frac{60±60\sqrt{2}}{50}$$

Ebatzi 25x^{2}-60x-36=0 ekuazioa ± plus denean eta ± minus denean.

$$x=\frac{6-6\sqrt{2}}{5} x=\frac{6\sqrt{2}+6}{5}$$

Zerrendatu aurkitutako ebazpen guztiak.

$$x=-\frac{6}{5} x=\frac{6-6\sqrt{2}}{5} x=\frac{6\sqrt{2}+6}{5}$$