Réitigh {l}{a^2+b^2=100}{a+b=20}

$$\left\{ \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 100 } \\ { a + b = 20 } \end{array} \right.$$

Réitigh do a,b.

$$a=10+5\sqrt{2}i\approx 10+7.071067812i\text{, }b=-5\sqrt{2}i+10\approx 10-7.071067812i$$

$$a=-5\sqrt{2}i+10\approx 10-7.071067812i\text{, }b=10+5\sqrt{2}i\approx 10+7.071067812i$$

Céimeanna ag Baint Úsáid as Ionadú agus an bhFoirme Chearnach

Réitigh a+b=20 do a trí a ar an taobh clé den chomhartha ‘Cothrom le’ a aonrú.

$$a+b=20$$

Bain b ón dá thaobh den chothromóid.

$$a=-b+20$$

Cuir a in aonad -b+20 sa chothromóid eile, b^{2}+a^{2}=100.

$$b^{2}+\left(-b+20\right)^{2}=100$$

Cearnóg -b+20.

$$b^{2}+b^{2}-40b+400=100$$

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

$$2b^{2}-40b+400=100$$

Bain 100 ón dá thaobh den chothromóid.

$$2b^{2}-40b+300=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir 1+1\left(-1\right)^{2} in ionad a, 1\times 20\left(-1\right)\times 2 in ionad b, agus 300 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 2\times 300}}{2\times 2}$$

Cearnóg 1\times 20\left(-1\right)\times 2.

$$b=\frac{-\left(-40\right)±\sqrt{1600-4\times 2\times 300}}{2\times 2}$$

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

$$b=\frac{-\left(-40\right)±\sqrt{1600-8\times 300}}{2\times 2}$$

Méadaigh -8 faoi 300.

$$b=\frac{-\left(-40\right)±\sqrt{1600-2400}}{2\times 2}$$

Suimigh 1600 le -2400?

$$b=\frac{-\left(-40\right)±\sqrt{-800}}{2\times 2}$$

Tóg fréamh chearnach -800.

$$b=\frac{-\left(-40\right)±20\sqrt{2}i}{2\times 2}$$

Tá 40 urchomhairleach le 1\times 20\left(-1\right)\times 2.

$$b=\frac{40±20\sqrt{2}i}{2\times 2}$$

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

$$b=\frac{40±20\sqrt{2}i}{4}$$

Réitigh an chothromóid b=\frac{40±20\sqrt{2}i}{4} nuair is ionann ± agus plus. Suimigh 40 le 20i\sqrt{2}?

$$b=\frac{40+20\sqrt{2}i}{4}$$

Roinn 40+20i\sqrt{2} faoi 4.

$$b=10+5\sqrt{2}i$$

Réitigh an chothromóid b=\frac{40±20\sqrt{2}i}{4} nuair is ionann ± agus míneas. Dealaigh 20i\sqrt{2} ó 40.

$$b=\frac{-20\sqrt{2}i+40}{4}$$

Roinn 40-20i\sqrt{2} faoi 4.

$$b=-5\sqrt{2}i+10$$

Tá dhá réiteach ann do b: 10+5i\sqrt{2} agus 10-5i\sqrt{2}. Cuir b in aonad 10+5i\sqrt{2} sa chothromóid eile a=-b+20 chun an réiteach comhfhreagrach do a a shásaíonn an dá chothromóid a fháil.

$$a=-\left(10+5\sqrt{2}i\right)+20$$

Ansin cuir b in aonad 10-5i\sqrt{2} sa chothromóid eile a=-b+20 agus faigh réiteach chun an réiteach comhfhreagrach do a a shásaíonn an dá chothromóid a fháil.

$$a=-\left(-5\sqrt{2}i+10\right)+20$$

Tá an córas réitithe anois.

$$a=-\left(10+5\sqrt{2}i\right)+20,b=10+5\sqrt{2}i\text{ or }a=-\left(-5\sqrt{2}i+10\right)+20,b=-5\sqrt{2}i+10$$