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Whakaoti mō x
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Tohaina

x^{2}-2x-5=0
Kia whakaotia te koreōrite, me tauwehe te taha mauī. Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-5\right)}}{2}
Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -2 mō te b, me te -5 mō te c i te ture pūrua.
x=\frac{2±2\sqrt{6}}{2}
Mahia ngā tātaitai.
x=\sqrt{6}+1 x=1-\sqrt{6}
Whakaotia te whārite x=\frac{2±2\sqrt{6}}{2} ina he tōrunga te ±, ina he tōraro te ±.
\left(x-\left(\sqrt{6}+1\right)\right)\left(x-\left(1-\sqrt{6}\right)\right)<0
Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.
x-\left(\sqrt{6}+1\right)>0 x-\left(1-\sqrt{6}\right)<0
Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\left(\sqrt{6}+1\right) me te x-\left(1-\sqrt{6}\right). Whakaarohia te tauira ina he tōrunga te x-\left(\sqrt{6}+1\right) he tōraro te x-\left(1-\sqrt{6}\right).
x\in \emptyset
He teka tēnei mō tētahi x ahakoa.
x-\left(1-\sqrt{6}\right)>0 x-\left(\sqrt{6}+1\right)<0
Whakaarohia te tauira ina he tōrunga te x-\left(1-\sqrt{6}\right) he tōraro te x-\left(\sqrt{6}+1\right).
x\in \left(1-\sqrt{6},\sqrt{6}+1\right)
Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(1-\sqrt{6},\sqrt{6}+1\right).
x\in \left(1-\sqrt{6},\sqrt{6}+1\right)
Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.