Whakaoti i te (x^2-3x+5)+(2x^2-7x-4)=

$$( x ^ { 2 } - 3 x + 5 ) + ( 2 x ^ { 2 } - 7 x - 4 ) =$$

Aromātai

$$3x^{2}-10x+1$$

Ngā Upane Otinga

Pahekotia te x^{2} me 2x^{2}, ka 3x^{2}.

$$3x^{2}-3x+5-7x-4$$

Pahekotia te -3x me -7x, ka -10x.

$$3x^{2}-10x+5-4$$

Tangohia te 4 i te 5, ka 1.

$$3x^{2}-10x+1$$

Tauwehe

$$3\left(x-\frac{5-\sqrt{22}}{3}\right)\left(x-\frac{\sqrt{22}+5}{3}\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Pahekotia te x^{2} me 2x^{2}, ka 3x^{2}.

$$factor(3x^{2}-3x+5-7x-4)$$

Pahekotia te -3x me -7x, ka -10x.

$$factor(3x^{2}-10x+5-4)$$

Tangohia te 4 i te 5, ka 1.

$$factor(3x^{2}-10x+1)$$

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.

$$3x^{2}-10x+1=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3}}{2\times 3}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\times 3}$$

Tāpiri 100 ki te -12.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\times 3}$$

Tuhia te pūtakerua o te 88.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\times 3}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±2\sqrt{22}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{10±2\sqrt{22}}{6}$$

Nā, me whakaoti te whārite x=\frac{10±2\sqrt{22}}{6} ina he tāpiri te ±. Tāpiri 10 ki te 2\sqrt{22}.

$$x=\frac{2\sqrt{22}+10}{6}$$

Whakawehe 10+2\sqrt{22} ki te 6.

$$x=\frac{\sqrt{22}+5}{3}$$

Nā, me whakaoti te whārite x=\frac{10±2\sqrt{22}}{6} ina he tango te ±. Tango 2\sqrt{22} mai i 10.

$$x=\frac{10-2\sqrt{22}}{6}$$

Whakawehe 10-2\sqrt{22} ki te 6.

$$x=\frac{5-\sqrt{22}}{3}$$

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+\sqrt{22}}{3} mō te x_{1} me te \frac{5-\sqrt{22}}{3} mō te x_{2}.

$$3x^{2}-10x+1=3\left(x-\frac{\sqrt{22}+5}{3}\right)\left(x-\frac{5-\sqrt{22}}{3}\right)$$