Whakaoti i te (x+3)-10-(x-2)^2=(x-1)^2-(x-5)^2+4+x

$$( x + 3 ) - 10 - ( x - 2 ) ^ { 2 } = ( x - 1 ) ^ { 2 } - ( x - 5 ) ^ { 2 } + 4 + x$$

Whakaoti mō x

$$x=\sqrt{13}-2\approx 1.605551275$$

$$x=-\left(\sqrt{13}+2\right)\approx -5.605551275$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 10 i te 3, ka -7.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Hei kimi i te tauaro o x^{2}-4x+4, kimihia te tauaro o ia taurangi.

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Pahekotia te x me 4x, ka 5x.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Tangohia te 4 i te -7, ka -11.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-5\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

Hei kimi i te tauaro o x^{2}-10x+25, kimihia te tauaro o ia taurangi.

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

Pahekotia te -2x me 10x, ka 8x.

$$5x-11-x^{2}=8x+1-25+4+x$$

Tangohia te 25 i te 1, ka -24.

$$5x-11-x^{2}=8x-24+4+x$$

Tāpirihia te -24 ki te 4, ka -20.

$$5x-11-x^{2}=8x-20+x$$

Pahekotia te 8x me x, ka 9x.

$$5x-11-x^{2}=9x-20$$

Tangohia te 9x mai i ngā taha e rua.

$$5x-11-x^{2}-9x=-20$$

Pahekotia te 5x me -9x, ka -4x.

$$-4x-11-x^{2}=-20$$

Me tāpiri te 20 ki ngā taha e rua.

$$-4x-11-x^{2}+20=0$$

Tāpirihia te -11 ki te 20, ka 9.

$$-4x+9-x^{2}=0$$

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.

$$-x^{2}-4x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, -4 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 9}}{2\left(-1\right)}$$

Whakareatia 4 ki te 9.

$$x=\frac{-\left(-4\right)±\sqrt{16+36}}{2\left(-1\right)}$$

Tāpiri 16 ki te 36.

$$x=\frac{-\left(-4\right)±\sqrt{52}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 52.

$$x=\frac{-\left(-4\right)±2\sqrt{13}}{2\left(-1\right)}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{13}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{4±2\sqrt{13}}{-2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{13}}{-2} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+4}{-2}$$

Whakawehe 4+2\sqrt{13} ki te -2.

$$x=-\left(\sqrt{13}+2\right)$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{13}}{-2} ina he tango te ±. Tango 2\sqrt{13} mai i 4.

$$x=\frac{4-2\sqrt{13}}{-2}$$

Whakawehe 4-2\sqrt{13} ki te -2.

$$x=\sqrt{13}-2$$

Kua oti te whārite te whakatau.

$$x=-\left(\sqrt{13}+2\right) x=\sqrt{13}-2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 10 i te 3, ka -7.

$$x-7-\left(x-2\right)^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

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

$$x-7-\left(x^{2}-4x+4\right)=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Hei kimi i te tauaro o x^{2}-4x+4, kimihia te tauaro o ia taurangi.

$$x-7-x^{2}+4x-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Pahekotia te x me 4x, ka 5x.

$$5x-7-x^{2}-4=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Tangohia te 4 i te -7, ka -11.

$$5x-11-x^{2}=\left(x-1\right)^{2}-\left(x-5\right)^{2}+4+x$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x-5\right)^{2}+4+x$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-5\right)^{2}.

$$5x-11-x^{2}=x^{2}-2x+1-\left(x^{2}-10x+25\right)+4+x$$

Hei kimi i te tauaro o x^{2}-10x+25, kimihia te tauaro o ia taurangi.

$$5x-11-x^{2}=x^{2}-2x+1-x^{2}+10x-25+4+x$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$5x-11-x^{2}=-2x+1+10x-25+4+x$$

Pahekotia te -2x me 10x, ka 8x.

$$5x-11-x^{2}=8x+1-25+4+x$$

Tangohia te 25 i te 1, ka -24.

$$5x-11-x^{2}=8x-24+4+x$$

Tāpirihia te -24 ki te 4, ka -20.

$$5x-11-x^{2}=8x-20+x$$

Pahekotia te 8x me x, ka 9x.

$$5x-11-x^{2}=9x-20$$

Tangohia te 9x mai i ngā taha e rua.

$$5x-11-x^{2}-9x=-20$$

Pahekotia te 5x me -9x, ka -4x.

$$-4x-11-x^{2}=-20$$

Me tāpiri te 11 ki ngā taha e rua.

$$-4x-x^{2}=-20+11$$

Tāpirihia te -20 ki te 11, ka -9.

$$-4x-x^{2}=-9$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-x^{2}-4x=-9$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{9}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{9}{-1}$$

Whakawehe -4 ki te -1.

$$x^{2}+4x=-\frac{9}{-1}$$

Whakawehe -9 ki te -1.

$$x^{2}+4x=9$$

Whakawehea te 4, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 2. Nā, tāpiria te pūrua o te 2 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+4x+2^{2}=9+2^{2}$$

Pūrua 2.

$$x^{2}+4x+4=9+4$$

Tāpiri 9 ki te 4.

$$x^{2}+4x+4=13$$

Tauwehea x^{2}+4x+4. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=13$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{13}$$

Whakarūnātia.

$$x+2=\sqrt{13} x+2=-\sqrt{13}$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$x=\sqrt{13}-2 x=-\sqrt{13}-2$$