Réitigh {l}{a+b=6}{a^2+b^2=6}

$$\left. \begin{array} { l } { a + b = 6 } \\ { a ^ { 2 } + b ^ { 2 } = 6 } \end{array} \right.$$

Réitigh do a,b.

$$a=3+\sqrt{6}i\approx 3+2.449489743i\text{, }b=-\sqrt{6}i+3\approx 3-2.449489743i$$

$$a=-\sqrt{6}i+3\approx 3-2.449489743i\text{, }b=3+\sqrt{6}i\approx 3+2.449489743i$$

Céimeanna ag Baint Úsáid as Ionadú agus an bhFoirme Chearnach

Réitigh a+b=6 do a trí a ar an taobh clé den chomhartha ‘Cothrom le’ a aonrú.

$$a+b=6$$

Bain b ón dá thaobh den chothromóid.

$$a=-b+6$$

Cuir a in aonad -b+6 sa chothromóid eile, b^{2}+a^{2}=6.

$$b^{2}+\left(-b+6\right)^{2}=6$$

Cearnóg -b+6.

$$b^{2}+b^{2}-12b+36=6$$

Suimigh b^{2} le b^{2}?

$$2b^{2}-12b+36=6$$

Bain 6 ón dá thaobh den chothromóid.

$$2b^{2}-12b+30=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir 1+1\left(-1\right)^{2} in ionad a, 1\times 6\left(-1\right)\times 2 in ionad b, agus 30 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 30}}{2\times 2}$$

Cearnóg 1\times 6\left(-1\right)\times 2.

$$b=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 30}}{2\times 2}$$

Méadaigh -4 faoi 1+1\left(-1\right)^{2}.

$$b=\frac{-\left(-12\right)±\sqrt{144-8\times 30}}{2\times 2}$$

Méadaigh -8 faoi 30.

$$b=\frac{-\left(-12\right)±\sqrt{144-240}}{2\times 2}$$

Suimigh 144 le -240?

$$b=\frac{-\left(-12\right)±\sqrt{-96}}{2\times 2}$$

Tóg fréamh chearnach -96.

$$b=\frac{-\left(-12\right)±4\sqrt{6}i}{2\times 2}$$

Tá 12 urchomhairleach le 1\times 6\left(-1\right)\times 2.

$$b=\frac{12±4\sqrt{6}i}{2\times 2}$$

Méadaigh 2 faoi 1+1\left(-1\right)^{2}.

$$b=\frac{12±4\sqrt{6}i}{4}$$

Réitigh an chothromóid b=\frac{12±4\sqrt{6}i}{4} nuair is ionann ± agus plus. Suimigh 12 le 4i\sqrt{6}?

$$b=\frac{12+4\sqrt{6}i}{4}$$

Roinn 12+4i\sqrt{6} faoi 4.

$$b=3+\sqrt{6}i$$

Réitigh an chothromóid b=\frac{12±4\sqrt{6}i}{4} nuair is ionann ± agus míneas. Dealaigh 4i\sqrt{6} ó 12.

$$b=\frac{-4\sqrt{6}i+12}{4}$$

Roinn 12-4i\sqrt{6} faoi 4.

$$b=-\sqrt{6}i+3$$

Tá dhá réiteach ann do b: 3+i\sqrt{6} agus 3-i\sqrt{6}. Cuir b in aonad 3+i\sqrt{6} sa chothromóid eile a=-b+6 chun an réiteach comhfhreagrach do a a shásaíonn an dá chothromóid a fháil.

$$a=-\left(3+\sqrt{6}i\right)+6$$

Ansin cuir b in aonad 3-i\sqrt{6} sa chothromóid eile a=-b+6 agus faigh réiteach chun an réiteach comhfhreagrach do a a shásaíonn an dá chothromóid a fháil.

$$a=-\left(-\sqrt{6}i+3\right)+6$$

Tá an córas réitithe anois.

$$a=-\left(3+\sqrt{6}i\right)+6,b=3+\sqrt{6}i\text{ or }a=-\left(-\sqrt{6}i+3\right)+6,b=-\sqrt{6}i+3$$