Whakaoti i te 4x^2+8x-45<0

$$4 x ^ { 2 } + 8 x - 45 < 0$$

Whakaoti mō x

$$x\in \left(-\frac{9}{2},\frac{5}{2}\right)$$

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.

$$4x^{2}+8x-45=0$$

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 4 mō te a, te 8 mō te b, me te -45 mō te c i te ture pūrua.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\left(-45\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{-8±28}{8}$$

Whakaotia te whārite x=\frac{-8±28}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{5}{2} x=-\frac{9}{2}$$

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

$$4\left(x-\frac{5}{2}\right)\left(x+\frac{9}{2}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{5}{2} me te x+\frac{9}{2}. Whakaarohia te tauira ina he tōrunga te x-\frac{5}{2} he tōraro te x+\frac{9}{2}.

$$x-\frac{5}{2}>0 x+\frac{9}{2}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+\frac{9}{2} he tōraro te x-\frac{5}{2}.

$$x+\frac{9}{2}>0 x-\frac{5}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-\frac{9}{2},\frac{5}{2}\right).

$$x\in \left(-\frac{9}{2},\frac{5}{2}\right)$$

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

$$x\in \left(-\frac{9}{2},\frac{5}{2}\right)$$