Whakaoti i te x^2-10x=13

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

Whakaoti mō x

$$x=\sqrt{38}+5\approx 11.164414003$$

$$x=5-\sqrt{38}\approx -1.164414003$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

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

Me tango 13 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 13 i a ia ake anō ka toe ko te 0.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -10 mō b, me -13 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-13\right)}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-13\right)}}{2}$$

Whakareatia -4 ki te -13.

$$x=\frac{-\left(-10\right)±\sqrt{100+52}}{2}$$

Tāpiri 100 ki te 52.

$$x=\frac{-\left(-10\right)±\sqrt{152}}{2}$$

Tuhia te pūtakerua o te 152.

$$x=\frac{-\left(-10\right)±2\sqrt{38}}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±2\sqrt{38}}{2}$$

Nā, me whakaoti te whārite x=\frac{10±2\sqrt{38}}{2} ina he tāpiri te ±. Tāpiri 10 ki te 2\sqrt{38}.

$$x=\frac{2\sqrt{38}+10}{2}$$

Whakawehe 10+2\sqrt{38} ki te 2.

$$x=\sqrt{38}+5$$

Nā, me whakaoti te whārite x=\frac{10±2\sqrt{38}}{2} ina he tango te ±. Tango 2\sqrt{38} mai i 10.

$$x=\frac{10-2\sqrt{38}}{2}$$

Whakawehe 10-2\sqrt{38} ki te 2.

$$x=5-\sqrt{38}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{38}+5 x=5-\sqrt{38}$$

Ngā Upane mō te Whakaoti i te Pūrua

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.

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

Whakawehea te -10, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -5. Nā, tāpiria te pūrua o te -5 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-10x+\left(-5\right)^{2}=13+\left(-5\right)^{2}$$

Pūrua -5.

$$x^{2}-10x+25=13+25$$

Tāpiri 13 ki te 25.

$$x^{2}-10x+25=38$$

Tauwehea x^{2}-10x+25. 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}.

$$\left(x-5\right)^{2}=38$$

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

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{38}$$

Whakarūnātia.

$$x-5=\sqrt{38} x-5=-\sqrt{38}$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$x=\sqrt{38}+5 x=5-\sqrt{38}$$