Whakaoti i te -3x^2+17x-20=

$$- 3 x ^ { 2 } + 17 x - 20 =$$

Tauwehe

$$\left(4-x\right)\left(3x-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -3x^{2}+ax+bx-20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=17 ab=-3\left(-20\right)=60$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Tātaihia te tapeke mō ia takirua.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

Ko te otinga te takirua ka hoatu i te tapeke 17.

$$a=12 b=5$$

Tuhia anō te -3x^{2}+17x-20 hei \left(-3x^{2}+12x\right)+\left(5x-20\right).

$$\left(-3x^{2}+12x\right)+\left(5x-20\right)$$

Tauwehea te 3x i te tuatahi me te -5 i te rōpū tuarua.

$$3x\left(-x+4\right)-5\left(-x+4\right)$$

Whakatauwehea atu te kīanga pātahi -x+4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+4\right)\left(3x-5\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$-3x^{2}+17x-20=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-3\right)\left(-20\right)}}{2\left(-3\right)}$$

Pūrua 17.

$$x=\frac{-17±\sqrt{289-4\left(-3\right)\left(-20\right)}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-17±\sqrt{289+12\left(-20\right)}}{2\left(-3\right)}$$

Whakareatia 12 ki te -20.

$$x=\frac{-17±\sqrt{289-240}}{2\left(-3\right)}$$

Tāpiri 289 ki te -240.

$$x=\frac{-17±\sqrt{49}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-17±7}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-17±7}{-6}$$

Nā, me whakaoti te whārite x=\frac{-17±7}{-6} ina he tāpiri te ±. Tāpiri -17 ki te 7.

$$x=-\frac{10}{-6}$$

Whakahekea te hautanga \frac{-10}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{5}{3}$$

Nā, me whakaoti te whārite x=\frac{-17±7}{-6} ina he tango te ±. Tango 7 mai i -17.

$$x=-\frac{24}{-6}$$

Whakawehe -24 ki te -6.

$$x=4$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te \frac{5}{3} mō te x_{1} me te 4 mō te x_{2}.

$$-3x^{2}+17x-20=-3\left(x-\frac{5}{3}\right)\left(x-4\right)$$

Tango \frac{5}{3} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$-3x^{2}+17x-20=-3\times \frac{-3x+5}{-3}\left(x-4\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te -3 me te 3.

$$-3x^{2}+17x-20=\left(-3x+5\right)\left(x-4\right)$$

Aromātai

$$\left(4-x\right)\left(3x-5\right)$$