Whakaoti i te 128(1+x)(1+x)=200

$$128(1+x)(1+x)=200$$

Whakaoti mō x

$$x=\frac{1}{4}=0.25$$

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 1+x ki te 1+x, ka \left(1+x\right)^{2}.

$$128\left(1+x\right)^{2}=200$$

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

$$128\left(1+2x+x^{2}\right)=200$$

Whakamahia te āhuatanga tohatoha hei whakarea te 128 ki te 1+2x+x^{2}.

$$128+256x+128x^{2}=200$$

Tangohia te 200 mai i ngā taha e rua.

$$128+256x+128x^{2}-200=0$$

Tangohia te 200 i te 128, ka -72.

$$-72+256x+128x^{2}=0$$

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.

$$128x^{2}+256x-72=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 128 mō a, 256 mō b, me -72 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-256±\sqrt{256^{2}-4\times 128\left(-72\right)}}{2\times 128}$$

Pūrua 256.

$$x=\frac{-256±\sqrt{65536-4\times 128\left(-72\right)}}{2\times 128}$$

Whakareatia -4 ki te 128.

$$x=\frac{-256±\sqrt{65536-512\left(-72\right)}}{2\times 128}$$

Whakareatia -512 ki te -72.

$$x=\frac{-256±\sqrt{65536+36864}}{2\times 128}$$

Tāpiri 65536 ki te 36864.

$$x=\frac{-256±\sqrt{102400}}{2\times 128}$$

Tuhia te pūtakerua o te 102400.

$$x=\frac{-256±320}{2\times 128}$$

Whakareatia 2 ki te 128.

$$x=\frac{-256±320}{256}$$

Nā, me whakaoti te whārite x=\frac{-256±320}{256} ina he tāpiri te ±. Tāpiri -256 ki te 320.

$$x=\frac{64}{256}$$

Whakahekea te hautanga \frac{64}{256} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 64.

$$x=\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{-256±320}{256} ina he tango te ±. Tango 320 mai i -256.

$$x=-\frac{576}{256}$$

Whakahekea te hautanga \frac{-576}{256} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 64.

$$x=-\frac{9}{4}$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 1+x ki te 1+x, ka \left(1+x\right)^{2}.

$$128\left(1+x\right)^{2}=200$$

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

$$128\left(1+2x+x^{2}\right)=200$$

Whakamahia te āhuatanga tohatoha hei whakarea te 128 ki te 1+2x+x^{2}.

$$128+256x+128x^{2}=200$$

Tangohia te 128 mai i ngā taha e rua.

$$256x+128x^{2}=200-128$$

Tangohia te 128 i te 200, ka 72.

$$256x+128x^{2}=72$$

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.

$$128x^{2}+256x=72$$

Whakawehea ngā taha e rua ki te 128.

$$\frac{128x^{2}+256x}{128}=\frac{72}{128}$$

Mā te whakawehe ki te 128 ka wetekia te whakareanga ki te 128.

$$x^{2}+\frac{256}{128}x=\frac{72}{128}$$

Whakawehe 256 ki te 128.

$$x^{2}+2x=\frac{72}{128}$$

Whakahekea te hautanga \frac{72}{128} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$x^{2}+2x=\frac{9}{16}$$

Whakawehea te 2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 1. Nā, tāpiria te pūrua o te 1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+2x+1^{2}=\frac{9}{16}+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=\frac{9}{16}+1$$

Tāpiri \frac{9}{16} ki te 1.

$$x^{2}+2x+1=\frac{25}{16}$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$x+1=\frac{5}{4} x+1=-\frac{5}{4}$$

Me tango 1 mai i ngā taha e rua o te whārite.

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