Whakaoti i te x^2-4x-5=2

$${ x }^{ 2 } -4x-5 = 2$$

Whakaoti mō x

$$x=\sqrt{11}+2\approx 5.31662479$$

$$x=2-\sqrt{11}\approx -1.31662479$$

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.

$$x^{2}-4x-5=2$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$x^{2}-4x-5-2=2-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$x^{2}-4x-5-2=0$$

Tango 2 mai i -5.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -4 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-7\right)}}{2}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-7\right)}}{2}$$

Whakareatia -4 ki te -7.

$$x=\frac{-\left(-4\right)±\sqrt{16+28}}{2}$$

Tāpiri 16 ki te 28.

$$x=\frac{-\left(-4\right)±\sqrt{44}}{2}$$

Tuhia te pūtakerua o te 44.

$$x=\frac{-\left(-4\right)±2\sqrt{11}}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{11}}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{11}}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+4}{2}$$

Whakawehe 4+2\sqrt{11} ki te 2.

$$x=\sqrt{11}+2$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{11}}{2} ina he tango te ±. Tango 2\sqrt{11} mai i 4.

$$x=\frac{4-2\sqrt{11}}{2}$$

Whakawehe 4-2\sqrt{11} ki te 2.

$$x=2-\sqrt{11}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{11}+2 x=2-\sqrt{11}$$

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.

$$x^{2}-4x-5=2$$

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

$$x^{2}-4x-5-\left(-5\right)=2-\left(-5\right)$$

Mā te tango i te -5 i a ia ake anō ka toe ko te 0.

$$x^{2}-4x=2-\left(-5\right)$$

Tango -5 mai i 2.

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

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

$$x^{2}-4x+\left(-2\right)^{2}=7+\left(-2\right)^{2}$$

Pūrua -2.

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

Tāpiri 7 ki te 4.

$$x^{2}-4x+4=11$$

Tauwehea x^{2}-4x+4. 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}.

$$\left(x-2\right)^{2}=11$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{11}$$

Whakarūnātia.

$$x-2=\sqrt{11} x-2=-\sqrt{11}$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=\sqrt{11}+2 x=2-\sqrt{11}$$