Whakaoti i te b^2+2b-47>0

$$b ^ { 2 } + 2 b - 47 > 0$$

Whakaoti mō b

$$b\in \left(-\infty,-4\sqrt{3}-1\right)\cup \left(4\sqrt{3}-1,\infty\right)$$

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.

$$b^{2}+2b-47=0$$

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 -47 mō te c i te ture pūrua.

$$b=\frac{-2±\sqrt{2^{2}-4\times 1\left(-47\right)}}{2}$$

Mahia ngā tātaitai.

$$b=\frac{-2±8\sqrt{3}}{2}$$

Whakaotia te whārite b=\frac{-2±8\sqrt{3}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$b=4\sqrt{3}-1 b=-4\sqrt{3}-1$$

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

$$\left(b-\left(4\sqrt{3}-1\right)\right)\left(b-\left(-4\sqrt{3}-1\right)\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te b-\left(4\sqrt{3}-1\right) me te b-\left(-4\sqrt{3}-1\right), me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te b-\left(4\sqrt{3}-1\right) me te b-\left(-4\sqrt{3}-1\right).

$$b-\left(4\sqrt{3}-1\right)<0 b-\left(-4\sqrt{3}-1\right)<0$$

Te otinga e whakaea i ngā koreōrite e rua ko b<-4\sqrt{3}-1.

$$b<-4\sqrt{3}-1$$

Whakaarohia te tauira ina he tōrunga tahi te b-\left(4\sqrt{3}-1\right) me te b-\left(-4\sqrt{3}-1\right).

$$b-\left(-4\sqrt{3}-1\right)>0 b-\left(4\sqrt{3}-1\right)>0$$

Te otinga e whakaea i ngā koreōrite e rua ko b>4\sqrt{3}-1.

$$b>4\sqrt{3}-1$$

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

$$b<-4\sqrt{3}-1\text{; }b>4\sqrt{3}-1$$