Whakaoti i te 4left(x+3right)^2-12x-72>0

$$4 { \left(x+3 \right) }^{ 2 } -12x-72 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,\frac{-3\sqrt{5}-3}{2}\right)\cup \left(\frac{3\sqrt{5}-3}{2},\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+3\right)^{2}.

$$4\left(x^{2}+6x+9\right)-12x-72>0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+6x+9.

$$4x^{2}+24x+36-12x-72>0$$

Pahekotia te 24x me -12x, ka 12x.

$$4x^{2}+12x+36-72>0$$

Tangohia te 72 i te 36, ka -36.

$$4x^{2}+12x-36>0$$

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.

$$4x^{2}+12x-36=0$$

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 12 mō te b, me te -36 mō te c i te ture pūrua.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{-12±12\sqrt{5}}{8}$$

Whakaotia te whārite x=\frac{-12±12\sqrt{5}}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{3\sqrt{5}-3}{2} x=\frac{-3\sqrt{5}-3}{2}$$

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

$$4\left(x-\frac{3\sqrt{5}-3}{2}\right)\left(x-\frac{-3\sqrt{5}-3}{2}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{3\sqrt{5}-3}{2} me te x-\frac{-3\sqrt{5}-3}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{3\sqrt{5}-3}{2} me te x-\frac{-3\sqrt{5}-3}{2}.

$$x-\frac{3\sqrt{5}-3}{2}<0 x-\frac{-3\sqrt{5}-3}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<\frac{-3\sqrt{5}-3}{2}.

$$x<\frac{-3\sqrt{5}-3}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{3\sqrt{5}-3}{2} me te x-\frac{-3\sqrt{5}-3}{2}.

$$x-\frac{-3\sqrt{5}-3}{2}>0 x-\frac{3\sqrt{5}-3}{2}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{3\sqrt{5}-3}{2}.

$$x>\frac{3\sqrt{5}-3}{2}$$

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

$$x<\frac{-3\sqrt{5}-3}{2}\text{; }x>\frac{3\sqrt{5}-3}{2}$$