Whakaoti i te x-4[x*2(x+6)]=5x+3

$$x - 4 [ x \cdot 2 ( x + 6 ) ] = 5 x + 3$$

Whakaoti mō x

$$x=\frac{\sqrt{163}-13}{4}\approx -0.058213666$$

$$x=\frac{-\sqrt{163}-13}{4}\approx -6.441786334$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 4 ki te 2, ka 8.

$$x-8x\left(x+6\right)=5x+3$$

Tangohia te 5x mai i ngā taha e rua.

$$x-8x\left(x+6\right)-5x=3$$

Tangohia te 3 mai i ngā taha e rua.

$$x-8x\left(x+6\right)-5x-3=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -8x ki te x+6.

$$x-8x^{2}-48x-5x-3=0$$

Pahekotia te x me -48x, ka -47x.

$$-47x-8x^{2}-5x-3=0$$

Pahekotia te -47x me -5x, ka -52x.

$$-52x-8x^{2}-3=0$$

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.

$$-8x^{2}-52x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -8 mō a, -52 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Pūrua -52.

$$x=\frac{-\left(-52\right)±\sqrt{2704-4\left(-8\right)\left(-3\right)}}{2\left(-8\right)}$$

Whakareatia -4 ki te -8.

$$x=\frac{-\left(-52\right)±\sqrt{2704+32\left(-3\right)}}{2\left(-8\right)}$$

Whakareatia 32 ki te -3.

$$x=\frac{-\left(-52\right)±\sqrt{2704-96}}{2\left(-8\right)}$$

Tāpiri 2704 ki te -96.

$$x=\frac{-\left(-52\right)±\sqrt{2608}}{2\left(-8\right)}$$

Tuhia te pūtakerua o te 2608.

$$x=\frac{-\left(-52\right)±4\sqrt{163}}{2\left(-8\right)}$$

Ko te tauaro o -52 ko 52.

$$x=\frac{52±4\sqrt{163}}{2\left(-8\right)}$$

Whakareatia 2 ki te -8.

$$x=\frac{52±4\sqrt{163}}{-16}$$

Nā, me whakaoti te whārite x=\frac{52±4\sqrt{163}}{-16} ina he tāpiri te ±. Tāpiri 52 ki te 4\sqrt{163}.

$$x=\frac{4\sqrt{163}+52}{-16}$$

Whakawehe 52+4\sqrt{163} ki te -16.

$$x=\frac{-\sqrt{163}-13}{4}$$

Nā, me whakaoti te whārite x=\frac{52±4\sqrt{163}}{-16} ina he tango te ±. Tango 4\sqrt{163} mai i 52.

$$x=\frac{52-4\sqrt{163}}{-16}$$

Whakawehe 52-4\sqrt{163} ki te -16.

$$x=\frac{\sqrt{163}-13}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{163}-13}{4} x=\frac{\sqrt{163}-13}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 4 ki te 2, ka 8.

$$x-8x\left(x+6\right)=5x+3$$

Tangohia te 5x mai i ngā taha e rua.

$$x-8x\left(x+6\right)-5x=3$$

Whakamahia te āhuatanga tohatoha hei whakarea te -8x ki te x+6.

$$x-8x^{2}-48x-5x=3$$

Pahekotia te x me -48x, ka -47x.

$$-47x-8x^{2}-5x=3$$

Pahekotia te -47x me -5x, ka -52x.

$$-52x-8x^{2}=3$$

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.

$$-8x^{2}-52x=3$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x^{2}-52x}{-8}=\frac{3}{-8}$$

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x^{2}+\left(-\frac{52}{-8}\right)x=\frac{3}{-8}$$

Whakahekea te hautanga \frac{-52}{-8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x^{2}+\frac{13}{2}x=\frac{3}{-8}$$

Whakawehe 3 ki te -8.

$$x^{2}+\frac{13}{2}x=-\frac{3}{8}$$

Whakawehea te \frac{13}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{13}{4}. Nā, tāpiria te pūrua o te \frac{13}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=-\frac{3}{8}+\left(\frac{13}{4}\right)^{2}$$

Pūruatia \frac{13}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=-\frac{3}{8}+\frac{169}{16}$$

Tāpiri -\frac{3}{8} ki te \frac{169}{16} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{163}{16}$$

Tauwehea x^{2}+\frac{13}{2}x+\frac{169}{16}. 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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{163}{16}$$

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

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{163}{16}}$$

Whakarūnātia.

$$x+\frac{13}{4}=\frac{\sqrt{163}}{4} x+\frac{13}{4}=-\frac{\sqrt{163}}{4}$$

Me tango \frac{13}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{163}-13}{4} x=\frac{-\sqrt{163}-13}{4}$$