Whakaoti i te 7x^2-5x+6>0

$$7 { x }^{ 2 } -5x+6 > 0$$

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.

$$7x^{2}-5x+6=0$$

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 7 mō te a, te -5 mō te b, me te 6 mō te c i te ture pūrua.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\times 6}}{2\times 7}$$

Mahia ngā tātaitai.

$$x=\frac{5±\sqrt{-143}}{14}$$

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 7x^{2}-5x+6 ki tā tētahi x. Kia whakatau i te tohu, tātaitia te uara o te kīanga mō x=0.

$$7\times 0^{2}-5\times 0+6=6$$

Ko te uara o te kīanga 7x^{2}-5x+6 he tōrunga i ngā wā katoa. E mau ana te koreōrite mō x\in \mathrm{R}.

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