Whakaoti i te {l}{m^2-2m}{-3<0}

$$\left. \begin{array} { l } { m ^ { 2 } - 2 m } \\ { - 3 < 0 } \end{array} \right.$$

Whakaoti mō m

$$m\in \left(-1,3\right)$$

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.

$$m^{2}-2m-3=0$$

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

$$m=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$m=\frac{2±4}{2}$$

Whakaotia te whārite m=\frac{2±4}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$m=3 m=-1$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$\left(m-3\right)\left(m+1\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te m-3 me te m+1. Whakaarohia te tauira ina he tōrunga te m-3 he tōraro te m+1.

$$m-3>0 m+1<0$$

He teka tēnei mō tētahi m ahakoa.

$$m\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te m+1 he tōraro te m-3.

$$m+1>0 m-3<0$$

Te otinga e whakaea i ngā koreōrite e rua ko m\in \left(-1,3\right).

$$m\in \left(-1,3\right)$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$m\in \left(-1,3\right)$$