Whakaoti i te x^2+x+1>0

$$x ^ { 2 } + x + 1 > 0$$

Whakaoti mō x

$$x\in \mathrm{R}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$x^{2}+x+1=0$$

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 1 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{-3}}{2}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā. He rite te tohu o te kīanga x^{2}+x+1 ki tā tētahi x. Kia whakatau i te tohu, tātaitia te uara o te kīanga mō x=0.

$$0^{2}+0+1=1$$

Ko te uara o te kīanga x^{2}+x+1 he tōrunga i ngā wā katoa. E mau ana te koreōrite mō x\in \mathrm{R}.

$$x\in \mathrm{R}$$