Whakaoti i te x^2-8x+10=13x

$$x ^ { 2 } - 8 x + 10 = 13 x$$

Whakaoti mō x

$$x = \frac{\sqrt{401} + 21}{2} \approx 20.512492197$$

$$x=\frac{21-\sqrt{401}}{2}\approx 0.487507803$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 13x mai i ngā taha e rua.

$$x^{2}-8x+10-13x=0$$

Pahekotia te -8x me -13x, ka -21x.

$$x^{2}-21x+10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -21 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 10}}{2}$$

Pūrua -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 10}}{2}$$

Whakareatia -4 ki te 10.

$$x=\frac{-\left(-21\right)±\sqrt{441-40}}{2}$$

Tāpiri 441 ki te -40.

$$x=\frac{-\left(-21\right)±\sqrt{401}}{2}$$

Ko te tauaro o -21 ko 21.

$$x=\frac{21±\sqrt{401}}{2}$$

Nā, me whakaoti te whārite x=\frac{21±\sqrt{401}}{2} ina he tāpiri te ±. Tāpiri 21 ki te \sqrt{401}.

$$x=\frac{\sqrt{401}+21}{2}$$

Nā, me whakaoti te whārite x=\frac{21±\sqrt{401}}{2} ina he tango te ±. Tango \sqrt{401} mai i 21.

$$x=\frac{21-\sqrt{401}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 13x mai i ngā taha e rua.

$$x^{2}-8x+10-13x=0$$

Pahekotia te -8x me -13x, ka -21x.

$$x^{2}-21x+10=0$$

Tangohia te 10 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-21x=-10$$

Whakawehea te -21, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{21}{2}. Nā, tāpiria te pūrua o te -\frac{21}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-21x+\left(-\frac{21}{2}\right)^{2}=-10+\left(-\frac{21}{2}\right)^{2}$$

Pūruatia -\frac{21}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-21x+\frac{441}{4}=-10+\frac{441}{4}$$

Tāpiri -10 ki te \frac{441}{4}.

$$x^{2}-21x+\frac{441}{4}=\frac{401}{4}$$

Tauwehea te x^{2}-21x+\frac{441}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{21}{2}\right)^{2}=\frac{401}{4}$$

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

$$\sqrt{\left(x-\frac{21}{2}\right)^{2}}=\sqrt{\frac{401}{4}}$$

Whakarūnātia.

$$x-\frac{21}{2}=\frac{\sqrt{401}}{2} x-\frac{21}{2}=-\frac{\sqrt{401}}{2}$$

Me tāpiri \frac{21}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$