Whakaoti i te 5x^2=40x+5

$$5 x ^ { 2 } = 40 x + 5$$

Whakaoti mō x

$$x=\sqrt{17}+4\approx 8.123105626$$

$$x=4-\sqrt{17}\approx -0.123105626$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 40x mai i ngā taha e rua.

$$5x^{2}-40x=5$$

Tangohia te 5 mai i ngā taha e rua.

$$5x^{2}-40x-5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -40 mō b, me -5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Pūrua -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-5\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-5\right)}}{2\times 5}$$

Whakareatia -20 ki te -5.

$$x=\frac{-\left(-40\right)±\sqrt{1600+100}}{2\times 5}$$

Tāpiri 1600 ki te 100.

$$x=\frac{-\left(-40\right)±\sqrt{1700}}{2\times 5}$$

Tuhia te pūtakerua o te 1700.

$$x=\frac{-\left(-40\right)±10\sqrt{17}}{2\times 5}$$

Ko te tauaro o -40 ko 40.

$$x=\frac{40±10\sqrt{17}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{40±10\sqrt{17}}{10}$$

Nā, me whakaoti te whārite x=\frac{40±10\sqrt{17}}{10} ina he tāpiri te ±. Tāpiri 40 ki te 10\sqrt{17}.

$$x=\frac{10\sqrt{17}+40}{10}$$

Whakawehe 40+10\sqrt{17} ki te 10.

$$x=\sqrt{17}+4$$

Nā, me whakaoti te whārite x=\frac{40±10\sqrt{17}}{10} ina he tango te ±. Tango 10\sqrt{17} mai i 40.

$$x=\frac{40-10\sqrt{17}}{10}$$

Whakawehe 40-10\sqrt{17} ki te 10.

$$x=4-\sqrt{17}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{17}+4 x=4-\sqrt{17}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 40x mai i ngā taha e rua.

$$5x^{2}-40x=5$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-40x}{5}=\frac{5}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}+\left(-\frac{40}{5}\right)x=\frac{5}{5}$$

Whakawehe -40 ki te 5.

$$x^{2}-8x=\frac{5}{5}$$

Whakawehe 5 ki te 5.

$$x^{2}-8x=1$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=1+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=1+16$$

Tāpiri 1 ki te 16.

$$x^{2}-8x+16=17$$

Tauwehea x^{2}-8x+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}.

$$\left(x-4\right)^{2}=17$$

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{17}$$

Whakarūnātia.

$$x-4=\sqrt{17} x-4=-\sqrt{17}$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=\sqrt{17}+4 x=4-\sqrt{17}$$