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

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

Whakaoti mō a

$$a=5$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

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

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

Tangohia te a^{2} mai i ngā taha e rua.

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

Pahekotia te a^{2} me -a^{2}, ka 0.

$$-4a+20=0$$

Tangohia te 20 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-4a=-20$$

Whakawehea ngā taha e rua ki te -4.

$$a=\frac{-20}{-4}$$

Whakawehea te -20 ki te -4, kia riro ko 5.

$$a=5$$

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

$$\sqrt{5^{2}-4\times 5+20}=5$$

Whakarūnātia. Ko te uara a=5 kua ngata te whārite.

$$5=5$$

Ko te whārite \sqrt{a^{2}-4a+20}=a he rongoā ahurei.

$$a=5$$