Réitigh x^2(6%)^2+(1-x)^2(2%)^2+2x(1-x)*012*6%*2%=00327

$$x ^ { 2 } ( 6 \% ) ^ { 2 } + ( 1 - x ) ^ { 2 } ( 2 \% ) ^ { 2 } + 2 x ( 1 - x ) \times 012 \times 6 \% \times 2 \% = 00327$$

Réitigh do x. (complex solution)

$$x=\frac{1}{10}+\frac{3}{10}i=0.1+0.3i$$

$$x=\frac{1}{10}-\frac{3}{10}i=0.1-0.3i$$

Céimeanna ag Baint Úsáid as an bhFoirmle Chearnach

Laghdaigh an codán \frac{6}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$x^{2}\times \left(\frac{3}{50}\right)^{2}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Ríomh cumhacht \frac{3}{50} de 2 agus faigh \frac{9}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(1-x\right)^{2} a leathnú.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{2}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{1}{50}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Ríomh cumhacht \frac{1}{50} de 2 agus faigh \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \frac{1}{2500}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Úsáid an t-airí dáileach chun 1-2x+x^{2} a mhéadú faoi \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\frac{1}{2500}-\frac{1}{1250}x+\frac{1}{2500}x^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Comhcheangail x^{2}\times \frac{9}{2500} agus \frac{1}{2500}x^{2} chun \frac{1}{250}x^{2} a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 2 agus 0 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 0 agus 12 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{6}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{3}{50}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 0 agus \frac{3}{50} chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{2}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{1}{50}=0\times 0\times 327$$

Méadaigh 0 agus \frac{1}{50} chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)=0\times 0\times 327$$

Is ionann rud ar bith méadaithe faoi nialas agus nialas.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0=0\times 0\times 327$$

Suimigh \frac{1}{2500} agus 0 chun \frac{1}{2500} a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 0\times 327$$

Méadaigh 0 agus 0 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 327$$

Méadaigh 0 agus 327 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0$$

Is féidir gach cothromóid san fhoirm ax^{2}+bx+c=0 a réiteach ag baint úsáid as an bhfoirmle chearnach : \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Tugann an fhoirmle chearnach dhá réiteach, ceann amháin nuair is suimiú é ± agus ceann eile nuair is dealú é.

$$\frac{1}{250}x^{2}-\frac{1}{1250}x+\frac{1}{2500}=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir \frac{1}{250} in ionad a, -\frac{1}{1250} in ionad b, agus \frac{1}{2500} in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\left(-\frac{1}{1250}\right)^{2}-4\times \frac{1}{250}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Cearnaigh -\frac{1}{1250} trí uimhreoir agus ainmneoir an chodáin a chearnú.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-4\times \frac{1}{250}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Méadaigh -4 faoi \frac{1}{250}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-\frac{2}{125}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Méadaigh -\frac{2}{125} faoi \frac{1}{2500} tríd an uimhreoir a mhéadú faoin uimhreoir agus an t-ainmneoir a mhéadú faoin ainmneoir. Laghdaigh an codán ansin go dtí na téarmaí is ísle más féidir.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-\frac{1}{156250}}}{2\times \frac{1}{250}}$$

Suimigh \frac{1}{1562500} le -\frac{1}{156250} trí chomhainmneoir a fháil agus na huimhreoirí a shuimiú. Laghdaigh an codán ansin go dtí na téarmaí is ísle más féidir.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{-\frac{9}{1562500}}}{2\times \frac{1}{250}}$$

Tóg fréamh chearnach -\frac{9}{1562500}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\frac{3}{1250}i}{2\times \frac{1}{250}}$$

Tá \frac{1}{1250} urchomhairleach le -\frac{1}{1250}.

$$x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{2\times \frac{1}{250}}$$

Méadaigh 2 faoi \frac{1}{250}.

$$x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}}$$

Réitigh an chothromóid x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}} nuair is ionann ± agus plus. Suimigh \frac{1}{1250} le \frac{3}{1250}i?

$$x=\frac{\frac{1}{1250}+\frac{3}{1250}i}{\frac{1}{125}}$$

Roinn \frac{1}{1250}+\frac{3}{1250}i faoi \frac{1}{125} trí \frac{1}{1250}+\frac{3}{1250}i a mhéadú faoi dheilín \frac{1}{125}.

$$x=\frac{1}{10}+\frac{3}{10}i$$

Réitigh an chothromóid x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}} nuair is ionann ± agus míneas. Dealaigh \frac{3}{1250}i ó \frac{1}{1250}.

$$x=\frac{\frac{1}{1250}-\frac{3}{1250}i}{\frac{1}{125}}$$

Roinn \frac{1}{1250}-\frac{3}{1250}i faoi \frac{1}{125} trí \frac{1}{1250}-\frac{3}{1250}i a mhéadú faoi dheilín \frac{1}{125}.

$$x=\frac{1}{10}-\frac{3}{10}i$$

Tá an chothromóid réitithe anois.

$$x=\frac{1}{10}+\frac{3}{10}i x=\frac{1}{10}-\frac{3}{10}i$$

Céimeanna chun an Chearnóg a Chomhlánú

Laghdaigh an codán \frac{6}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$x^{2}\times \left(\frac{3}{50}\right)^{2}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Ríomh cumhacht \frac{3}{50} de 2 agus faigh \frac{9}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(1-x\right)^{2} a leathnú.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{2}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{1}{50}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Ríomh cumhacht \frac{1}{50} de 2 agus faigh \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \frac{1}{2500}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Úsáid an t-airí dáileach chun 1-2x+x^{2} a mhéadú faoi \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\frac{1}{2500}-\frac{1}{1250}x+\frac{1}{2500}x^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Comhcheangail x^{2}\times \frac{9}{2500} agus \frac{1}{2500}x^{2} chun \frac{1}{250}x^{2} a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 2 agus 0 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 0 agus 12 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{6}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{3}{50}\times \frac{2}{100}=0\times 0\times 327$$

Méadaigh 0 agus \frac{3}{50} chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{2}{100}=0\times 0\times 327$$

Laghdaigh an codán \frac{2}{100} chuig na téarmaí is ísle trí 2 a bhaint agus a chealú.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{1}{50}=0\times 0\times 327$$

Méadaigh 0 agus \frac{1}{50} chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)=0\times 0\times 327$$

Is ionann rud ar bith méadaithe faoi nialas agus nialas.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0=0\times 0\times 327$$

Suimigh \frac{1}{2500} agus 0 chun \frac{1}{2500} a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 0\times 327$$

Méadaigh 0 agus 0 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 327$$

Méadaigh 0 agus 327 chun 0 a fháil.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0$$

Bain \frac{1}{2500} ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

$$\frac{1}{250}x^{2}-\frac{1}{1250}x=-\frac{1}{2500}$$

Iolraigh an dá thaobh faoi 250.

$$\frac{\frac{1}{250}x^{2}-\frac{1}{1250}x}{\frac{1}{250}}=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Má roinntear é faoi \frac{1}{250} cuirtear an iolrúchán faoi \frac{1}{250} ar ceal.

$$x^{2}+\left(-\frac{\frac{1}{1250}}{\frac{1}{250}}\right)x=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Roinn -\frac{1}{1250} faoi \frac{1}{250} trí -\frac{1}{1250} a mhéadú faoi dheilín \frac{1}{250}.

$$x^{2}-\frac{1}{5}x=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Roinn -\frac{1}{2500} faoi \frac{1}{250} trí -\frac{1}{2500} a mhéadú faoi dheilín \frac{1}{250}.

$$x^{2}-\frac{1}{5}x=-\frac{1}{10}$$

Roinn -\frac{1}{5}, comhéifeacht an téarma x, faoi 2 chun -\frac{1}{10} a fháil. Ansin suimigh uimhir chearnach -\frac{1}{10} leis an dá thaobh den chothromóid. Déanann an chéim seo slánchearnóg de thaobh clé na cothromóide.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=-\frac{1}{10}+\left(-\frac{1}{10}\right)^{2}$$

Cearnaigh -\frac{1}{10} trí uimhreoir agus ainmneoir an chodáin a chearnú.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=-\frac{1}{10}+\frac{1}{100}$$

Suimigh -\frac{1}{10} le \frac{1}{100} trí chomhainmneoir a fháil agus na huimhreoirí a shuimiú. Laghdaigh an codán ansin go dtí na téarmaí is ísle más féidir.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=-\frac{9}{100}$$

Fachtóirigh x^{2}-\frac{1}{5}x+\frac{1}{100}. Go ginearálta, nuair x^{2}+bx+c cearnóg fhoirfe é, is féidir é a fhachtóiriú i gcónaí mar \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=-\frac{9}{100}$$

Tóg fréamh chearnach an dá thaobh den chothromóid.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{-\frac{9}{100}}$$

Simpligh.

$$x-\frac{1}{10}=\frac{3}{10}i x-\frac{1}{10}=-\frac{3}{10}i$$

Cuir \frac{1}{10} leis an dá thaobh den chothromóid.

$$x=\frac{1}{10}+\frac{3}{10}i x=\frac{1}{10}-\frac{3}{10}i$$