Whakaoti i te -2x^2-3x-10leq0

$$- 2 x ^ { 2 } - 3 x - 10 \leq 0$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -2x^{2}-3x-10. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$2x^{2}+3x+10\geq 0$$

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.

$$2x^{2}+3x+10=0$$

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 2 mō te a, te 3 mō te b, me te 10 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 10}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{-71}}{4}$$

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 2x^{2}+3x+10 ki tā tētahi x. Kia whakatau i te tohu, tātaitia te uara o te kīanga mō x=0.

$$2\times 0^{2}+3\times 0+10=10$$

Ko te uara o te kīanga 2x^{2}+3x+10 he tōrunga i ngā wā katoa. E mau ana te koreōrite mō x\in \mathrm{R}.

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