Whakaoti i te 60-x^2+4x>0

$$60- { x }^{ 2 } +4x > 0$$

Whakaoti mō x

$$x\in \left(-6,10\right)$$

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 60-x^{2}+4x. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$-60+x^{2}-4x<0$$

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.

$$-60+x^{2}-4x=0$$

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 -4 mō te b, me te -60 mō te c i te ture pūrua.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-60\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{4±16}{2}$$

Whakaotia te whārite x=\frac{4±16}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=10 x=-6$$

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

$$\left(x-10\right)\left(x+6\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-10 me te x+6. Whakaarohia te tauira ina he tōrunga te x-10 he tōraro te x+6.

$$x-10>0 x+6<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+6 he tōraro te x-10.

$$x+6>0 x-10<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-6,10\right).

$$x\in \left(-6,10\right)$$

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

$$x\in \left(-6,10\right)$$