Whakaoti i te 4x^2+7x-8=x

$$4 x ^ { 2 } + 7 x - 8 = x$$

Whakaoti mō x

$$x=\frac{\sqrt{41}-3}{4}\approx 0.850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2.350781059$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x mai i ngā taha e rua.

$$4x^{2}+7x-8-x=0$$

Pahekotia te 7x me -x, ka 6x.

$$4x^{2}+6x-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 6 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-8\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-6±\sqrt{36-16\left(-8\right)}}{2\times 4}$$

Whakareatia -16 ki te -8.

$$x=\frac{-6±\sqrt{36+128}}{2\times 4}$$

Tāpiri 36 ki te 128.

$$x=\frac{-6±\sqrt{164}}{2\times 4}$$

Tuhia te pūtakerua o te 164.

$$x=\frac{-6±2\sqrt{41}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-6±2\sqrt{41}}{8}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{41}}{8} ina he tāpiri te ±. Tāpiri -6 ki te 2\sqrt{41}.

$$x=\frac{2\sqrt{41}-6}{8}$$

Whakawehe -6+2\sqrt{41} ki te 8.

$$x=\frac{\sqrt{41}-3}{4}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{41}}{8} ina he tango te ±. Tango 2\sqrt{41} mai i -6.

$$x=\frac{-2\sqrt{41}-6}{8}$$

Whakawehe -6-2\sqrt{41} ki te 8.

$$x=\frac{-\sqrt{41}-3}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x mai i ngā taha e rua.

$$4x^{2}+7x-8-x=0$$

Pahekotia te 7x me -x, ka 6x.

$$4x^{2}+6x-8=0$$

Me tāpiri te 8 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$4x^{2}+6x=8$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+6x}{4}=\frac{8}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x^{2}+\frac{6}{4}x=\frac{8}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{3}{2}x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$x^{2}+\frac{3}{2}x=2$$

Whakawehea te \frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{4}. Nā, tāpiria te pūrua o te \frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=2+\frac{9}{16}$$

Tāpiri 2 ki te \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

Tauwehea x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

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

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Whakarūnātia.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$