Whakaoti i te sqrt{a^2-4a+20}=sqrt{a}

$$\sqrt { a ^ { 2 } - 4 a + 20 } = \sqrt { a }$$

Whakaoti mō a

$$a=2\sqrt{5}e^{\arctan(\frac{\sqrt{55}}{5})i}\approx 2.5+3.708099244i$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{a^{2}-4a+20}\right)^{2}=\left(\sqrt{a}\right)^{2}$$

Tātaihia te \sqrt{a^{2}-4a+20} mā te pū o 2, kia riro ko a^{2}-4a+20.

$$a^{2}-4a+20=\left(\sqrt{a}\right)^{2}$$

Tātaihia te \sqrt{a} mā te pū o 2, kia riro ko a.

$$a^{2}-4a+20=a$$

Tangohia te a mai i ngā taha e rua.

$$a^{2}-4a+20-a=0$$

Pahekotia te -4a me -a, ka -5a.

$$a^{2}-5a+20=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 20 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 20}}{2}$$

Pūrua -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 20}}{2}$$

Whakareatia -4 ki te 20.

$$a=\frac{-\left(-5\right)±\sqrt{25-80}}{2}$$

Tāpiri 25 ki te -80.

$$a=\frac{-\left(-5\right)±\sqrt{-55}}{2}$$

Tuhia te pūtakerua o te -55.

$$a=\frac{-\left(-5\right)±\sqrt{55}i}{2}$$

Ko te tauaro o -5 ko 5.

$$a=\frac{5±\sqrt{55}i}{2}$$

Nā, me whakaoti te whārite a=\frac{5±\sqrt{55}i}{2} ina he tāpiri te ±. Tāpiri 5 ki te i\sqrt{55}.

$$a=\frac{5+\sqrt{55}i}{2}$$

Nā, me whakaoti te whārite a=\frac{5±\sqrt{55}i}{2} ina he tango te ±. Tango i\sqrt{55} mai i 5.

$$a=\frac{-\sqrt{55}i+5}{2}$$

Kua oti te whārite te whakatau.

$$a=\frac{5+\sqrt{55}i}{2} a=\frac{-\sqrt{55}i+5}{2}$$

Whakakapia te \frac{5+\sqrt{55}i}{2} mō te a i te whārite \sqrt{a^{2}-4a+20}=\sqrt{a}.

$$\sqrt{\left(\frac{5+\sqrt{55}i}{2}\right)^{2}-4\times \frac{5+\sqrt{55}i}{2}+20}=\sqrt{\frac{5+\sqrt{55}i}{2}}$$

Whakarūnātia. Ko te uara a=\frac{5+\sqrt{55}i}{2} kua ngata te whārite.

$$\frac{1}{2}\left(10+2i\times 55^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(\frac{5}{2}+\frac{1}{2}i\times 55^{\frac{1}{2}}\right)^{\frac{1}{2}}$$

Whakakapia te \frac{-\sqrt{55}i+5}{2} mō te a i te whārite \sqrt{a^{2}-4a+20}=\sqrt{a}.

$$\sqrt{\left(\frac{-\sqrt{55}i+5}{2}\right)^{2}-4\times \frac{-\sqrt{55}i+5}{2}+20}=\sqrt{\frac{-\sqrt{55}i+5}{2}}$$

Whakarūnātia. Ko te uara a=\frac{-\sqrt{55}i+5}{2} kua ngata te whārite.

$$\frac{1}{2}\left(10-2i\times 55^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(-\frac{1}{2}i\times 55^{\frac{1}{2}}+\frac{5}{2}\right)^{\frac{1}{2}}$$

Rārangihia ngā rongoā katoa o \sqrt{a^{2}-4a+20}=\sqrt{a}.

$$a=\frac{5+\sqrt{55}i}{2} a=\frac{-\sqrt{55}i+5}{2}$$